Question

An object is thrown vertically upwards with velocity of 49 m/s calculate the hight attained by the object & also time required to attend in the height if the acceleration is 9.8m/s square​

Answer 1

Explanation:

Consider a formula,

2gH = v2 – u2

2 × (- 9.8) × H = 0 – (49)2

– 19.6 H = – 2401

H = 122.5 m

Now consider a formula,

v = u + g × t

0 = 49 + (- 9.8) × t

– 49 = – 9.8t

t = 5 sec

(1) The maximum height to which the ball rises = 122.5 m

(2) The total time ball takes to return to the surface of the earth = 5 + 5 = 10 sec.