A ball is thrown directly down from the top of a building with a speed of 12 m/s,and it lands on the ground with a velocity 3 times the initial velocity.How high is the building (in meters)?
A)12
B)59
C)234
D)4320
E)8760

Question

Quizplus · Accepted Answer

Let h be the height of the building and t be the time it takes for the ball to hit the ground. 
Using the equation of motion: v = u + gt, where v = final velocity, u = initial velocity, g = acceleration due to gravity and t = time, we can find the time taken for the ball to hit the ground:
3u = v = u + gt 
2u = gt 
t = 2u/g = 2(12)/9.8 = 2.45 s

Using the equation of motion: h = ut + (1/2)gt^2, we can find the height of the building:
h = ut + (1/2)gt^2 
= 12(2.45) + (1/2)(9.8)(2.45)^2 
= 29.4 + 29.4 
= 58.8 m ≈ 59 m 
Therefore, the answer is B) 59.

A Ball Is Thrown Directly Down from the Top of a Building