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vaieri [72.5K]
4 years ago
5

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

rows the softball with a velocity of
23.5
m/s
at an angle of
33.5
∘
above the horizontal. When the softball leaves her hand, it is
16.5
m
above the water. How far does the softball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.
Physics
1 answer:
Andre45 [30]4 years ago
4 0

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

  • A uniform motion along the horizontal (x) direction
  • A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

s=u_y t + \frac{1}{2}at^2

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

u_y is the initial vertical velocity of the ball, which is given by

u_y = u sin \theta

where

u = 23.5 m/s is the initial velocity

\theta=33.5^{\circ} is the angle of projection

Substituting,

u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s

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

Substituting everything into the equation we get:

-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s

Therefore, the horizontal distance covered in a time t is

d=v_x t

And substituting t = 3.59 s, we find

d=(19.6)(3.59)=70.4 m

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4
denpristay [2]
Answer= 8m/s

Because total Momentum before= total momentum after

Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48

Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u


Mb=Ma
48=6u
u=8m/s
3 0
3 years ago
1. (15 points) Small cart is rolling down an inclined track and accelerating. It fires a ball straight out of the cannon as it m
Elena L [17]

Answer:

Explanation:

This problem can be solved with the conservation of the momentum.

If the ball is fired upward, the momentum before and after the ball is fired must conserve. Hence, the speed of the ball is the same that the speed of the car just in the moment in wich the ball is fired.

Hence, the result depends of the acceleration of the car. If the change in the speed is higher than the speed of the ball, it is probably that the ball will be behind the car or it will come back to the car.

If the ball is fired forward, and if the change in the speed of the car is not enogh, the ball will be in front of the car.

HOPE THIS HELPS!!

5 0
3 years ago
What does the area under a speed-time graph represent​
Tom [10]

Answer: It represents the whole distance traveled. Hope this helps!

Explanation:

4 0
3 years ago
In trampoline competitions, a good jump is one that lasts about 1.8 seconds. (A) How high can an athlete who stays in the air 1.
KonstantinChe [14]

Answer:

3.97305 m

Explanation:

a = Acceleration due to gravity = 9.81 m/s²

If a jump lasts for 1.8 seconds this means that from the moment when the person leaves the ground till the person touches the ground again it takes 1.8 seconds. So, maximum height reached will be at half the time of the jump i.e., 0.9 seconds.

u = Initial velocity = 0

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow s=0t+\frac{1}{2}9.81\times 0.9^2\\\Rightarrow s=3.97305\ m

So, height of the jump is 3.97305 m.

4 0
4 years ago
Calculate the percent error of the distanc
Soloha48 [4]

Answer:

5.25%

Explanation:

From the question given above, the following data were obtained:

Accepted value = 238857 miles

Measured value = 226316 miles

Percentage error =.?

Next, we shall determine the absolute error. This can be obtained as follow:

Accepted value = 238857 miles

Measured value = 226316 miles

Absolute Error =?

Absolute Error = |Measured – Accepted|

Absolute Error = |226316 – 238857|

Absolute Error = 12541

Finally, we shall determine the percentage error. This can be obtained as follow:

Accepted value = 238857 miles

Absolute Error = 12541

Percentage error =.?

Percentage error = absolute error / accepted value × 100

Percentage error

= 12541 / 238857 × 100

= 1254100 / 238857

= 5.25%

Therefore, the percentage error is 5.25%.

8 0
3 years ago
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