Question

A ball is thrown upward. After reaching a
maximum height, it continues falling back toward
Earth. On the way down, the ball is
caught at the same height at which it was
thrown upward.
v0
hmax
If the time (up and down) the ball remains
in the air is 1.84 s, find its speed
when it caught. The acceleration of gravity
is 9.8 m/s
2
. Neglect air resistance.
Answer in units of m/s.

Answer 1

GIVEN :

TOTAL TIME OF FLIGHT = 1.84 S

g = 9.8 m/s^2

SOLUTION :

Let V0 be the initial speed by which ball is thrown

after reaching maximum height,

final velocity = 0 m/s

as total time in air is 1.84

so, time taken in going up

 = 1.84/2

  = 0.92 s

As we know that

v = u + a t

so, 0 = V0 +(-9.8)(0.92)

so, V0 = 9.016 m/s

ALSO , s = u t + 1/2 a t^2

Let hmax be the maximum height

so, s = hmax = 9.016 + 1/2 (-9.8)(0.92)^2

or, hmax = 9.016 - 4.14736 m

hmax = 4.86864 m

As the ball is caught at same height

the velocity will be the same as V0 9.016 m/s

because the time taken to reach

back will be same as hmax and g is fixed and

also velocity at top is fixed as 0m/s.

Answer 2

Heya...

Kindly refer to above attachment...