A ball is thrown straight up from the edge of the roof of a building. A second b

Question

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.16s later. You may ignore air resistance.

a)If the height of the building is 19.1 m , what must the initial speed be of the first ball if both are to hit the ground at the same time?

b) Consider the same situation, but now let the initial speed v_0 of the first ball be given and treat the height 'h' of the building as an unknown. What must the height of the building be for both balls to reach the ground at the same time for v_0 = 8.75 m/s.

c) If v_0 is greater than some value v_max , a value of h does not exist that allows both balls to hit the ground at the same time. Solve for v_max .

d) If v_0 is greater than some value v_max, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for v_max .

Explanation / Answer

nswer

follow this

Acceleration of gravity is 9.8 m/s

Ball continues to rise for (50/9.8) seconds. Then its speed is zero before it starts to fall.

During the ascent, average speed is 1/2 (50 + 0) = 25 m/s.

Distance straight up = (average speed) x (time of ascent) = 25 x (50/9.8) = 127.55 meters(rounded).

This is one heck of a toss ! Almost 420 feet up ... like the height of a 40-story building ... before it starts to fall.

But 50 meters per second is a launch at 109 mph ( ! ), so I guess I'll stand by my math.

Note that as always, air resistance has been ignored

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