Two students are on a balcony a distance h above the street. One student throws

Question

Two students are on a balcony a distance h above the street. One student throws a ball vertically downward at a speed vi; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi, g, h, and t. (Take upward to be the positive direction.) (a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground? t = (b) Find the velocity of each ball as it strikes the ground. For the ball thrown upward vf = For the ball thrown downward vf = (c) How far apart are the balls at a time t after they are thrown and before they strike the ground? d =

Explanation / Answer

a) let t1 is time taken in the first case and t2 is the time taken in the second case

in the first case,

-h = -vi*t1 - 0.5*g*t1^2 ----(1)

in the second case

-h = vi*t2 - 0.5*g*t2^2 ---(2)

equation(2) - equation(1)

-h - (-h) = vi*t2 - 0.5*g*t2^2 - (-vi*t1 - 0.5*g*t1^2)

0 = vi*(t2 + t1) - 0.5*g*(t2^2 - t1^2)

vi*(t2 + t1) = 0.5*g*(t2^2 - t1^2)

vi*(t2 + t1) = 0.5*g*(t2 + t1)*(t2 - t1)

vi = 0.5*g*(t2 - t1)

t2 - t1 = 2*vi/g

delta_t = 2*vi/g <<<<<<<-------------Answer

b) for the ball through upward,

vf^2 - vi^2 = 2*(-g)*(-h)

vf = sqrt(vi^2 + 2*g*h)

for the ball through downward

vf^2 - vi^2 = 2*(-g)*(-h)

vf = sqrt(vi^2 + 2*g*h)

c)

d1 = vi*t - 0.5*g*t^2

d2 = vi*t + 0.5*t^2

d = d1 + d2

= 2*vi*t <<<<----------Answer

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