Ask question

Ask question

All categories

Varvara68 [4.7K]

2 years ago

10

Calculate the height of a cliff if it takes 2.35s for a rock to hit the ground when it is thrown straight up from the cliff with

a initial velocity 8m/s.

1 answer:

ad-work [718]2 years ago

7 0

Answer:

y₀ = 10.625 m

Explanation:

For this exercise we will use the kinematic relations, where the upward direction is positive.

y = y₀ + v₀ t - ½ g t²

in the exercise they indicate the initial velocity v₀ = 8 m / s.

when the rock reaches the ground its height is zero

0 = y₀ + v₀ t - ½ g t²

y₀i = -v₀ t + ½ g t²

let's calculate

y₀ = - 8 2.5 + ½ 9.8 2.5²

y₀ = 10.625 m

You might be interested in

The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha

Marizza181 [45]

Answer:

idk, idk cause i'm steppin on my toes and i can't stop i make flips ou of my flops

Explanation:

8 0

3 years ago

5. Find the mass of a car that is traveling at a velocity of 35 m/s West.

sattari [20]

Answer:

m = 9795.9 kg

Explanation:

v = 35 m/s

KE = 6,000,000 J

Plug those values into the following equation:

$KE = \frac{1}{2} mv^{2}$

6,000,000 J = (1/2)(35^2)m

---> m = 9795.9 kg

3 0

3 years ago

Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th

Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

$\theta=30^{\circ}$

So, the ball's horizontal velocity is

$v_x = u cos \theta = (15)(cos 30)=13.0 m/s$

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

$y=u_yt+\frac{1}{2}at^2$

where

$u_y = u sin \theta = (15)(sin 30) = 7.5 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

$2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0$

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

$d=v_x t =(13.0)(1.19)=15.5 m$

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

$v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s$

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

$d=98 m$

while the horizontal velocity of the ball is

$v_x = u cos \theta = (34)(cos 30)=29.4 m/s$

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

$t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s$

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

$y= h + u_y t + \frac{1}{2}at^2$

where

h = 1.2 m is the initial height

$u_y = u sin \theta = (34)(sin 30)=17.0 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting t = 3.33 s,

$y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m$

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

$u_y' = 7 m/s$

So its position of the glove at time t' is

$y'= h' + u_y' t' + \frac{1}{2}at'^2$

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

$3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0$

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

$t'' = t -t'$

So we have two solutions:

$t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s$

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

approximation to the average velocity in that time interval, what should be the sequence of calculations?Update the (vector) pos

Klio2033 [76]

Answer:

The steps are outlined in the explanation below.

Explanation:

The average velocity is derived midpoint from the initial to the final velocity. Here is the proof:

Find the total displacement:

let the displacement be given by the letter s

Then since the average velocity is defined as: $v_{av} = \frac{x - x_{0} }{t - t_{0} }$

where t = final time

t₀ = initial time

v = final speed

v₀ = initial time

where x denotes the position, then

$v_{ave} = \frac{v+v_{0} }{2}$

where v = $\frac{dx}{dt}$ and dx = change in distance with respect to time.

6 0

3 years ago

Light from a 560 nm monochromatic source is incident upon the surface of fused quartz (n=1. 56) at an angle of 60°. what is the

ankoles [38]

The angle of reflection is "60°".

Here we apply the law of the concept of reflection then we get the final answer easily.

The angle of incident = angle of reflection

Then, the Angle of the incident =60°

What is reflection?

Reflection is the phenomenon of light rays returning to the source after striking an obstruction.
It resembles the way a ball bounces when we toss it on a hard surface.
Some of the light rays that strike an item are reflected, some of them travel through it, and the remainder are absorbed by the object.
The given values are:Light from a monochromatic source,= 560 nm
The angle of incidence,= 60°
The surface of fused quartz (n),= 1.56
When a light ray does exist on a flat surface, the law or idea of reflection should apply since it includes both the reflected and "normal" light rays at the mirror surface.
According to the above law,Angle of incident = angle of reflection
Then, Angle of incident =60°.

To learn more about reflection visit: brainly.com/question/15487308

#SPJ4

7 0

2 years ago