Question

A ball is thrown off the top of a 75 ft high building. If it strikes the ground 60ft away from the building 3 seconds later, determine the initial velocity, the angle at launch and the magnitude of the ball's velocity when it hits the ground

Answer 1

Given that,

v = 75 ft = 22.86 m

x = 60 ft = 18.28 m

t = 3 s

(a)

Let, initial velocity = u

angle at launch =

Horizontal distance, x = ucos * t

18.28 = ucos * 3

ucos = 6.096 .........(1)

Vertical dstance, h = usin*t - (1/2)gt^2

-22.86 + (1/2)*9.8*3^2 =  usin * 3

21.24 =  usin * 3

usin​ = 7.08 .........(2)

From eq(1) and (2),

( ucos)^2 + (usin​ )^2 = (6.096)^2 + (7.08)^2

u = 9.34 m/s

(b)

From eq (1),

9.34 * cos = 6.096

= 49.3 deg

(c)

maximum height reached by ball

v^2 = uy^2 + 2gh

at maximum height, v = 0

0 = (7.08)^2 - 2*9.8*h

h = 2.55 m

Total distance H = 22.86 + 2.55 = 25.41 m

Now again from third equation of motion,

Here, velocity at highest = initial velocity = 0

so, v^2 - u^2 = 2gh

v^2 - 0 = 2*9.8*25.41

v = 22.32 m/s