A ball is thrown upward from the top of a 23.5 m tallbuilding. The ball\'s initi

Question

A ball is thrown upward from the top of a 23.5 m tallbuilding. The ball's initial speed is 11.2 m/s. At the sameinstant, a person is running on the ground at a distance of 31.6 mfrom the building. What must be the average speed of the person ifhe is to catch the ball at the bottom of the building?
Ive tried using the constant acceleration formula Vf- Vo / a =at, but to no avail. How would we account for the fact thatthe ball is being thrown from a certain height? Is that animportant factor in solving the problem? A ball is thrown upward from the top of a 23.5 m tallbuilding. The ball's initial speed is 11.2 m/s. At the sameinstant, a person is running on the ground at a distance of 31.6 mfrom the building. What must be the average speed of the person ifhe is to catch the ball at the bottom of the building?
Ive tried using the constant acceleration formula Vf- Vo / a =at, but to no avail. How would we account for the fact thatthe ball is being thrown from a certain height? Is that animportant factor in solving the problem?

Explanation / Answer

s = v0t + 1/2 a t2 If we take the upwards as positive then s = -23.5, g = -9.8,v0 = 11.2 4.9 t2 - 11.2 t + 23.5 = 0 Solving for t gives t = 3.61 sec S = V t   to get time for runner to reachbuilding If we take the upwards as positive then s = -23.5, g = -9.8,v0 = 11.2 4.9 t2 - 11.2 t + 23.5 = 0 Solving for t gives t = 3.61 sec S = V t   to get time for runner to reachbuilding

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