Problem 3. A ball is thrown straight up from an initial position yi = 1.00 m abo

Question

Problem 3.
A ball is thrown straight up from an initial position yi = 1.00 m above ground level with
an initial velocity v0. It reaches its top position after 2.00 s, and then drops down to the
ground. The value of free fall acceleration is g = 9.80 m/s2.
(a) What is the condition on the velocity at the top position of the ball?
(b) Determine the initial velocity of the ball.
(c) Calculate the height of the ball above the ground when it is at its highest point.
(d) What is the total time the ball spends in the air?

Explanation / Answer

a) velocity vy = 0

b) vy = voy - gt

0 = voy - 9.8*2

voy = 19.6 m/s

c) (yf-yi) = voy*t - 0.5*g*t^2

yf - 1 = (19.6*2)-(0.5*9.8*2^2)

yf - 1 = 19.6 m

yf = 20.6 m

d) as the ball reaches ground

0-1 = voy*t - 0.5*g*t^2

-1 = 19.6t - 4.9t^2

4.9t^2 - 19.6t - 1 = 0

t = 4.05s

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