Question

an object is thrown straight down from the top of a building at a speed of 20m/s. It hits the ground with a speed of,40m/s (g=9.8m/s square ) a,How high is the building b,How long was the object in the air​

Answer 1

Answer:

a)61.22 m

b)2.04 sec

Explanation:

a) v^2 - u^2 = 2*a*h

40^2 - 20^2 = 2*9.8 * h

1200/2* 9.8 = h

h = 61.22 m

b) v = u +a*t

t = (40-20)/9.8

t=2.04 sec.

Answer 2

Answer:

The height of the building is 61.22 m and the object was in the air for 2.04 seconds.

Explanation:

Initial speed = 20m/s

Final speed = 40m/s

[Given : g = 9.8m/s²]

Now, we can use laws of motion :

v² -  u² = 2 .g .s ----------------------(1)

Here,

v = final velocity

u  = initial velocity

g = acceleration due to gravity

Substituting values in eq (1) we get :

40² -  20² = 2.9.8. s

=> s = $\frac{1200}{19.6}$ = 61.22 m

v = u + gt  ---------------------(2)

=>  40 = 20 +9.8*t

=> t = 20/9.8 = 2.04 seconds

Therefore, the height of the building is 61.22 m and the object was in the air for 2.04 seconds.