Question

a body thrown vertically up with initial velocity 52 m/s from the ground twice a point at h height above at an interval of 10 s.find h

Answer 1

$here \: is \: \: your \: answer \: looking \: for$
initial velocity= 52 m/s (u)

interval of 10 s.

h =?

》using the formula ,

$s = ut + \frac{1}{2} g {t}^{2}$

》take g= 10

$s = 520 + \frac{1}{2} ( - 10)(10 {)}^{2}$
s = 520 − (5)(10)^2

s = 520 − 500 = 20 m

☆It is Given that,

=s = 2H

2H = 20 m

⇒H = 10 m.

$hope \: it \: helps$

$\beta e \: \beta \gamma \alpha inly$

Answer 2

$<b>Your answer !!$

● Given :

Initial Velocity = u = 52 m/s

Time = 10 sec

Find : h ?

● $<b>Using 2nd equation of motion <__>$

$= > s = ut + \frac{1}{2} a {t}^{2} \\ \\ acceleration \: due \: to \: gravitation \\ \\ = > s = ut + \frac{1}{2} g {t}^{2} \\ \\ = > s = 52 \times 10 + \frac{1}{2} \times - 9.8 \times {(10)}^{2} \\ \\ = > s = 520 + \frac{1}{2} \times \frac{- 98}{10} \times 100 \\ \\ = > s = 520 - 49 \times 10 \\ \\ = > 520 - 490 \\ \\ => s = 30 m$

● $<b>A.T.Q. According to Question !!$

$= > s = 2h \\ \\ = > 30 = 2h \\ \\ = > \frac{30}{2} = h \\ \\ = > 15 = h$

$<b>Thanks !!$