Question

A ball is thrown from a height of 10 feet at 45 degrees from the vertical with an initial speed of 40 ft/sec. Neglecting resistance and assuming the gravitational acceleration g=32 ft/sec^2 , determine:

a. Where is the ball after 2 seconds?

b. What is the velocity of the ball at 2 seconds, and which direction (up or down) the ball is going?

c. How high does the ball go?

Answer 1

a)

if the ball thrown down, then time of flight is given by 2nd equation of motion

h = ho + u cos 45 t + 0.5 gt^2

10 =40 * cos 45 t + 16.2 t^2

solving for t

t = 0.302 s

as the question asking for 2 s, then ball must be thrown upward.

position after 2 s is given by

y = 10 + 40 coa 45 * 2 — 0.5* 32.2* 2*2

y = 2.1685 ft

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b)

v = 40 cos 45 - 32.2* 2 = - 36.1 ft/s

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c)

max height from ground

h = 10 + ( 40 sin 45)^2/ 2g

h = 22.422 m

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