1)A baseball player standing on the ground throws a ball straight up. The ball l

Question

1)A baseball player standing on the ground throws a ball straight up. The ball leaves the player's hand with a speed of 3.13 m/s and is in the air for 2.10 s before hitting the ground. How high above the ground was the baseball player's hand when he released the ball?

2) A small child is climbing up a slide angled at 23.0° above the horizontal. They only make it 0.900 m up the slide before they give up, stop, and then slide back down. When the child reaches the bottom of the slide how fast will they be going?

Explanation / Answer

1)Given,

vi = 3.13 m/s ; t = 2.1 s

We know that,

Hmax = vi^2/2 g

Hmax = 3.13^2/2 x 9.81 = 0.49 m

time to travel this distance

t = sqrt (2 x 0.49/9.81) = 0.32 s

T = 2t = 0.32 s = 0.64 s

T' = 2.1 - 0.64 = 1.46 s

H = 0.5 x 9.8 x 1.46^2 = 10.44 m

Hence, his hand is 10.44 m high.

2)theta = 23 deg ; d = 0.9 m

h = d sin(theta)

h = 0.9 x sin23 = 0.352 m

from conservation of energy

1/2 m v^2 = m g h

v = sqrt (2 g h)

v = sqrt (2 x 9.8 x 0.352) = 2.63 m/s

Hence, v = 2.63 m/s

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