61.5k views
4 votes
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.

User Supi
by
8.3k points

Related questions

asked Dec 16, 2020 164k views
Wulfram asked Dec 16, 2020
by Wulfram
7.7k points
2 answers
0 votes
164k views