1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mixer [17]
3 years ago
11

A projectile is launched at ground level with an initial speed of 53.0 m/s at an angle of 35.0° above the horizontal. It strikes

a target above the ground 2.50 seconds later. What are the x and y distances from where the projectile was launched to where it lands?
Physics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

x = 108.5 m      y = 45.37 m,  

Explanation:

This is a projectile launching exercise, the horizontal distance it reaches is called the range and can be found by the expression

           x = v₀ₓ t

           

the initial hairiness is

          v₀ₓ = v₀ cos θ

           

           x = v₀ cos θ  t

  let's calculate

           x = 53 cos 35 2.50

           x = 108.5 m

the height of the projectile can be calculated with

            y =v₀ t - ½ g t²

             y = vo sin tea t - ½ g t²

            y = 53 sin 35 2.50 - ½ 9.8 2.5²

            y = 45.37 m

You might be interested in
A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is
pychu [463]

Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

v = u + at

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

7 0
3 years ago
WILL GIVE BRAINLIEST.
Tanzania [10]

The product of (mass) x (acceleration) is equal to (force).

So if (force) doesn't change and (mass) changes, then
(acceleration) has to change in the other direction from (mass)
in order to keep their product constant.

If the (forces) on both wagons are equal but one wagon has
double the (mass) of the other, then the more massive wagon
has  (1/double)  =  1/2  the acceleration of the less massive one.
3 0
3 years ago
Which material would you expect to have greater resistance, plastic or silver? Explain your answer.
olganol [36]

Answer:

plastic

Explanation:

because it is an insulator that it is a poor conductor of electricity.

8 0
3 years ago
A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altidude
Serjik [45]

Answer:

a) 252 ft/s

b) 1076 ft

Explanation:

The equation for motion for uniform acceleration (we can use it because the rocket is affected only by gravity) is as follows:

Y(t) = Y0 + V0 * t + 1/2 * a * t^2

Where

Y(t): altitude at a given time

Y0: initial altitude

V0: initial speed

a: acceleration, in this case -32.2 ft/s^2 (negative because gravity points down)

We set a 1 dimensional coordinate system with Y pointing up and the origin of coordinates at ground level.

We consider t=0 as the moment where powered flight ended (motor ran oou of fuel), at this moment the altitude was

Y(0) = 89.6 ft

Therefore:

Y0 = 89.6 ft

We also know that the rocket fell to ground 16 seconds later, therefore

Y(16) = 0 ft

So we can write

Y(t=16) = Y0 + V0 * t + 1/2 * a * t^2

V0 * t = Y(t=16) - Y0 - 1/2 * a * t^2

V0 =( Y(t=16) - Y0 - 1/2 * a * t^2 )/t

V0 =( 0 - 89.6 - 1/2 * (-32.2) * 16^2 )/16 = 252 ft/s

In the highest point of flight the rocket will have a speed = 0

The first derivative of the equation of motion is the equation of speed:

V(t) = V0 + a * t

If we equate this to zero we eill find the time at which the rocket achieved it's highest altitude.

0 = V0 + a * t

a * t = 0 - V0

t = -V0/a

t = -252/(-32.2) = 7.83 s

Now, we can take this time value andd plug it back into the position equation

Y(7.83) = 89.6 + 252 * 7.83 + 1/2 * (-32.2) * 7.83^2 = 1076 ft

6 0
3 years ago
A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the
Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

U=\dfrac{1}{2}kx^2...(I)

Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_{i}=E_{f}

U_{i}+U'_{i}=U_{f}+U'_{f}

\dfrac{1}{2}kx^2+0=0+mgh

h=\dfrac{kx^2}{2mg}

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}

h=11.653\ m

Hence, The maximum height above the point of release is 11.653 m.

3 0
4 years ago
Other questions:
  • Two point charges are located on the y-axis as follows: charge ????1 = −1.50 nC at y = −0.60 m, and charge ????2 = +3.20 nC at t
    8·1 answer
  • a student is pushing a 50 kilogram cart with a force of 500 newtons another students measures the speed of the cart and finds th
    8·1 answer
  • The Sun appears at an angle of 48.4° above the horizontal as viewed by a dolphin swimming underwater. What angle does the sunlig
    8·1 answer
  • A Person whose weight is 5.20 x 10^2 N is being pulledup
    6·1 answer
  • Which is an example of a solution?
    6·1 answer
  • What does someone with high self esteem look like
    10·1 answer
  • Three people push a piano on wheels with forces of 130 N to the right, 150 N to the left, and 165 N to the right. What is the st
    10·1 answer
  • Please correct answers only, if you don’t know, ignore. Please answer this or am I correct?
    15·2 answers
  • Why is the answer A and not C?<br>please help!​
    9·1 answer
  • What is an exoplanet?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!