Question

10) A ball is thrown up in the air with an

initial speed of 25 m/s. After a certain amount
of time, the ball is FALLING with a speed of 15
m/s. What is the average velocity of the ball at
that point?

a)-10 m/s
b) -5 m/s
c) + 10 m/s
d) +5 m/s

Answer 1

H = u²/(2g)

H = 25²/(2×10)

H = 31.25 m

The ball will reach 31.25 m high from the point of projection.

It reaches 30 + 31.25 = 61.25m high from the ground.

t1 = 2u/g

t1 = 2×25/10

t1 = 5 seconds

If v is the velocity with which it hits the ground then,

v = √(u² + 2aS)

v = √(25² + 2×10×30)

v = 35 m/s

v = u + at

t2 = (v - u)/g

t2 = (35 - 25)/10

t2 = 1 second

Time taken for it to hit the ground = t1 + t2

= 5 + 1

= 6 secondsSolution

H = u²/(2g)

H = 25²/(2×10)

H = 31.25 m

The ball will reach 31.25 m high from the point of projection.

It reaches 30 + 31.25 = 61.25m high from the ground.

t1 = 2u/g

t1 = 2×25/10

t1 = 5 seconds

If v is the velocity with which it hits the ground then,

v = √(u² + 2aS)

v = √(25² + 2×10×30)

v = 35 m/s

v = u + at

t2 = (v - u)/g

t2 = (35 - 25)/10

t2 = 1 second

Time taken for it to hit the ground = t1 + t2

= 5 + 1

= 6 seconds