Question

1. A ball is thrown upward at a speed of 26.0 m/s.

a) How long does it take before the ball reaches its maximum height (note, use 9.8 m/s(squared) as the magnitude of the acceleration due to gravity)

b) If it was thrown from a building, and it took 15.35s to hit the ground, how tall is the building?

c) What is the velocity just before it hits the ground?

Answer 1

1.

At the max height speed of ball, V = 0

U = Initial velocity = 26.0 m/sec

a = acceleration due to gravity = -g = -9.81 m/sec^2

So Using 1st kinematic equation:

V = U + a*t

t = (V - U)/a

t = (0 - 26)/(-9.81) = 2.65 sec

Time to reach max height = 2.65 sec

Part B.

total time taken by ball to reach ground = T = 15.35 sec

U = 26 sec

Suppose height of building = -H (Since height travelled is downward)

So Using 2nd kinematic equation:

d = U*t + (1/2)*a*t^2

-H = U*T + (1/2)*(-g)*T^2

H = -26*15.35 + (1/2)*9.81*15.35^2

H = 756.62 m = Height of building

Part C.

Using 1st kinematic equation:

V = U + a*t

V = 26.0 - (-9.81)*15.35

V = -124.58 m/sec (-ve sign means velocity is downward)

Speed = |V| = 124.58 m/sec

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