Question

4. Both trains speed up all the time.
5. Somewhere before time to both trains
have the same acceleration.
019 (part 1 of 3) 10.0 points
A tennis ball is dropped from 1.85 m above the
ground. It rebounds to a height of 0.947 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s. (Let
down be negative.)
Answer in units of m/s.
020 (part 2 of 3) 10.0 points
With what velocity does it leave the ground?
Answer in units of m/s.

Answer 1

Answer:

The velocity with which the ball strikes the ground = -5.7 m/s

The velocity with which the ball leaves the ground = 4.3 m/s

Explanation:

Seems like there's parts of a couple questions. I'll focus on the tennis ball part which seems to be a full question.

To find the velocity with which the tennis ball hits the ground, we only need to worry about what happens up to that point. We can ignore the rebound for this part. Given:

d = -1.85

a = -9.8

vi = 0

vf = ?

$v_f^2 = v_i^2 + 2ad\\v_f^2=0+2(-9.8)(-1.65)\\v_f^2=32.34\\v_f=-5.7$

*Keep in mind that the square root gives us two answers, a positve and a negative one. We use the negative one here because the final speed is downwards and the question says down is negative.

To find the velocity with which the ball leaves the ground, we only need to worry about what happens after that point. We can ignore the drop for this part. Given:

d = 0.947

a = -9.8

vf = 0

vi = ?

$v_f^2 = v_i^2 + 2ad\\0=v_i^2+2(-9.8)(0.947)\\v_i^2=18.56\\v_i=4.3$

*Here we use the positive speed after the square root because we know the ball is going up, which the question designates as the positive direction.