Question

John throws the ball straight upward and after 1 second it reaches maximum height then it does free fall motion which takes 2 seconds calculate the maximum height and velocity of ball before it crashes the ground (g=10m/s^

Answer 1

Answer:

When the ball goes up :-

t = 1 second

v= 0

g = - 10m/s^2----(since it is going in the opposite direction of acceleration , we take the value of g as negative)

u = ?

h = ?

We have , v = u + at ------(Newton's first equation of motion)

Therefore, v = u + gt

Therefore, 0 = u + (-10) × 1

Therefore, 0 = u - 10

Therefore, u = 10m/s

We have , s = ut + 1/2gt^2------( Newton's second equation of motion)

Therefore, s = 10×1 + 1/2 (-10)×1^2

Therefore, s = 10+1/2 × (-10)

Therefore, s= 10-5

Therefore, s= 5m

When the ball is falling down,

t = 2seconds

u = 0

g = 10m/s^2

v = ?

We Have, v = u + gt

Therefore, v = 0 + 10×2

Therefore, v = 20m/s

Answer :- The maximum height of the ball will be 5metres and it's velocity before crashing the ground will be 20 m/s.