Question

A ball is thrown vertically upwards IT returns 6 second later calculate the greatest height reached by the ball and initial velocity of the ball taking G is equal to 10 M per second square

Answer 1

A ball is thrown vertically upwards and it returns 6 second later.

So time of flight = 6s
time it took to reach maximum height = 6/2 = 3s

acceleration = -10 m/s^2 (downward is negative)

let initial velocity= u
final velocity at top , v = 0

we know that v = u + at

=> 0 = u - 10×3

=> 0 = u - 30

=> u = 30 m/s


Let the highest point reached = h meter

h = u^2 / 2g = 30^2 / (2×10) = 45 meter


Greatest height reached = 45m
initial velocity = 30 m/s

Answer 2

Total time taken for upward and downward journey  = 6 seconds

Time taken for downward journey = 3 seconds

For downward journey,

u = 0

t = 3 sec

g = 10 m/s^2

Since h = ut + 1/2 gt^2

h = 0 * 3 + 1/2 * 10 * 3^2

h = 0 + 5 * 9

h = 0 + 45 = 45 m

So,

Greatest height reached by the ball = 45 m

For upward journey,

v = 0

h = 45 m

g = 10 m/s^2

SInce, v^2 = u^2 - 2gh

0^2 = u^2 - 2 * 10 * 45

0 = u^2 - 900

u^2 = 900

u = $\bold{\sqrt{900}}$

u = 30 m/s

Initial velocity of the ball = 30 m/s

Thanking You!