Question

A ball rolls of a desk a speed of 3.0m/s and lands 0.40 seconds later

Answer 1

a) The horizontal distance covered by the ball is 1.2 m

b) The height of the desk is 0.784 m

c) The speed at the time of impact is 4.9 m/s

Explanation:

a)

We have to calculate how far from the base of the desk the ball lands, that is the horizontal distance covered by the ball during its motion.

The motion of the ball is a projectile motion and consists of two independent motions:

- A horizontal uniform motion at constant velocity, since no forces act along the horizontal direction

- A vertical accelerated motion with constant acceleration $g=9.8 m/s^2$ towards the ground, due to the force of gravity

In this part we are only interested in the horizontal motion. The horizontal velocity of the ball is constant and it is

$v_x = 3.0 m/s$

And the time of flight is

t = 0.40 s

Therefore, the horizontal distance covered is

$d=v_x t = (3.0)(0.40)=1.2 m$

b)

To find the height of the desk, we have to consider the vertical motion, which is a uniformly accelerated motion. Therefore, we can use the following suvat equation:

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

where:

s is the vertical displacement (= the height of the desk)

$u_y=0$ is the initial vertical velocity

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

t = 0.40 s is the time of flight

Substituting,

$s=0+\frac{1}{2}(9.8)(0.40)^2=0.784 m$

c)

To find the speed of the ball at the impact, we have to calculate both the horizontal and vertical components of the velocity at the time of impact.

We already said that the horizontal component is constant:

$v_x = 3.0 m/s$

While the vertical component is given by

$v_y = u_y + at$

where

$u_y = 0\\a=g=9.8 m/s^2\\t=0.40 s$

So

$v_y = 0+(9.8)(0.40)=3.9 m/s$

So, the speed of the ball at the time of impact is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(3.0)^2+(3.9)^2}=4.9 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

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