Question

plzzz solve the problem

Answer 1

We know that a certain height is reached two times in a vertical projection.(One while going upward,one while going back downward)

let the height reached in time t be h

maximum height=H

difference in height=H-h

now H is reached in T and h is reached in t time.

So H-h is reached in (T-t) time ...(1)

while going back down H-h is reached in (T-t)seconds

So,

H-h=u(T-t) + ½g(T-t)2

but u =0 (velocity at max height is zero)

So,

H-h = ½g(T-t)2

h = H – ½g(T-t)2 or H-½g(t-T)

HOPE IT HELP...
☣AMAN☣

Answer 2

Hey mate here is the correct answer...

consider the motion in vertical direction

v₀ = initial velocity

v = final velocity at maximum height = 0

H = maximum height gained

T = time taken to reach maximum height

a = acceleration = - g

Using the equation

v = v₀ + a t

0 = v₀ + (- g) T

v₀ = gT

height at any time is given as

Y = v₀ t + (0.5) a t²

Y = gTt + (0.5) (- g) t²

I hope it will be helpful to you...

Please mark me as BRAINLIEST ❤❤❤.....