Question

A boy throws a stone upward with a velocity of 60m/s.(a) How long will it take to reach the maximum height (g = -10m/s2)?(b) What is the maximum height reached by the ball?(c) How long will it take to reach the ground?

Answer 1

Given in the question :

➝ u = 60 m/s

➝ v = 0 m/s

➝ g = – 10 m/s²

Finding time taken by the stone to reach the maximum height using 1st equation of motion :

➝ v = u + gt

➝ 0 = 60 + (– 10) t

➝ 0 = 60 – 10t

➝ – 60 = – 10t

➝ – 60/– 10 = t

➝ t = 6 secs

Now, finding the maximum height reached by the ball using third equation of motion :

➝ v² – u² = 2gs

➝ (0)² – (60)² = 2 (– 10) s

➝ 0 – 360 = – 20s

➝ – 360/– 20 = s

➝ s = 180 m

Lastly, finding how long the ball will take to reach the ground using second equation of motion :

➝ s = ut + ½ gt²

Let the total time taken by the ball be T :

➝ T = t + t'

In the second equation of motion :

➝ s = displacement : – 180 m

➝ t' = time taken by it to reach ground

➝ u = initial velocity of it : 0 m/s

Substituting and solving for t' :

➝ s = ut' + ½ gt'²

➝ – 180 = (0) t' + ½ (– 10) t'²

➝ – 180 = – 5t'²

➝ – 180/– 5 = t'²

➝ 36 = t'²

➝ √36 = t'

➝ t' = 6 secs

Finding T by adding the values :

➝ T = t + t'

➝ T = 6 secs + 6 secs

➝ T = 12 seconds

________________________________

Therefore :

➝ ( a ) The stone will take 6 seconds to reach the maximum height.

➝ ( b ) The maximum height reached by the stone is 180 metres.

➝ ( c ) The stone will take 12 seconds to reach the ground.