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A ball is thrown upward from a tower 234 ft. high with an initial velocity of 19 ft per second. In how

many seconds will it strike the groun kod and with what velocity?​

1 Answer

7 votes

Explanation:

Set the origin of the coordinate system at the top of the tower.

Given:

y = -234 ft (when the ball hits the ground)

v0 = +19ft/a

g = 32 ft/s^2

a) The equation of motion. for this freely-falling body is

y = v0t - (1/2)gt^2

When the ball hits the ground, t = T (time of flight)

----> -234 ft = (19 ft/s)T - (1/2)(32 ft/s^2)T^2

Rearranging the terms and dropping the units momentarily for brevity, we get

(16)T^2 - 19T - 234 = 0

This is a quadratic equation whose solution is

T = (1/32)[19 +- {(-19)^2 - 4(16)(-234)}]^(1/2)

= 4.46 s. or -2.68 s

Therefore, the time of flight T = 4.46 s

b) To find the velocity upon impact, we use

v = v0 - gt, where t = T = 4.46 s

= (19 ft/s) - (32 ft/s^2)(4.46 s)

= -123.7 ft/s

The negative sign means that its velocity is pointing down.

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