Question

An object of 1 kg is thrown vertically upward with the velocity of 49 m/s.calculate

Also Find:
Maximum height attained by the object?
Time taken to reach the ground?

Answer 1

g = (v- u)/t
-9.8 = (0 - 49)/t
t = -49/(-9.8)
t = 5 sec

h( s) = ut -( 1/2) gt^2 ( since we are throwing the object against the gravitation force)
h = 49*5 - (1/2)*9.8 *5*5
h = 245- 4.9*25
h = 245- 122.5
h = 122.5 m
b ) time taken to reach maximum height will be equal to time taken to each the ground so
total time = time taken to reach maximum height + time taken to reach the ground = 5+5=10 sec