an individual is standing at the edge of a cliff. the person throws ball 1 upwar

Question

an individual is standing at the edge of a cliff. the person throws ball 1 upward with a velocity of 30 meters per second (m/s) and ball 2 downward at 30 m/s. assume there is no air resistance and that gravitational acceleration, g, is equal to 10m/s2. answer the following:

a. how long does it take ball 1 to reach its maximum height?
b. how high up does ball 1 go before turning and coming down?
c. what is the velocity of ball 1 at the top of its trajectory?
d. what is the velocity of ball 1 one second before reaching the top of its trajectory?
e. which ball will have the greater speed when it strikes the ground below?

Explanation / Answer

A. Initial velocity is 30 m/s. Acceleration is 10m/s^2 in the downward direction. The amount of time it will take to reach the peak is vf=vo+.5at2.

0=30+.5(-10)t2

t=2.45 seconds

B. sf=si+vo*t+.5at2

sf=30*2.45+.5*10*(2.45)2

103.51 meters

C. The final velocity is zero because the peak trajectory is when it stops.

D.vf=vo+.5at2

vf=30+.5(-10)*1.452

vf=19.49 m/s

E. The speed of ball 1 and ball 2 are going to be the same when they hit the ground because they both are traveling 30 m/s downward at the top of the cliff.

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