You throw a stone straight down from the top of a tall tower. It leaves your hand moving at 8.00 m/s, Air resistance can be neglected. Take the positive -direction to be upward, and choose y 0 to be the point where the stone leaves your hand. Find the stone's position 1.50s after it leaves your hand. Express your answer with the appropriate units.
Find the y-component of the stone's velocity 1.50 s after it leaves your hand. Express your answer with t0he appropriate units.

The velocity after 1.5 s is 22.7 m/s downwards.

Explanation:

Initial velocity = - 8 m/s

acceleration, a = - 9.8 m/s2

time, t = 1.5 s

Use first equation of motion

v = u + at

v = - 8 - 9.8 x 1.5

v = - 8 - 14.7

v = - 22.7 m/s  

Thus, the velocity after 1.5 s is 22.7 m/s downwards.

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