Question

A body is thrown up with a velocity 50m/s. Calculate the maximum height reached and the time taken to reach the maximum height ( a = 10 metre per second square​

Answer 1

ANSWER

u=50m/s

g=−10m/s

2

The velocity at the highest point will be zero

Thus, v=0

Height can be calculated as,

V-U= 2gs

0−2500=−20s

s=125m

Time can be calculated as,

v=u+gt

0=50+(−10)t

t=5sec

these is your answer thanks have a nice and great day.

Answer 2

Given :

• InitiaI veIocity of body = 50 m/s
• AcceIeration (a) = 10 m/s²

To find :

• Maximum height reached
• Time taken to reach maximum height

SoIution :

At maximum height, finaI veIocity of body is 0 m/s (v)

Now as body is thrown up so acceIeration due to gravity wiII be in negative i.e. -10 m/s² (g)

Now using 3rd equation of motion :

⇔ v² = u² + 2gh

⇒ 0² = (50)² + 2(-10) × h

⇒ 0 = 2500 - 20h

⇒ -2500 = -20h

⇒ h = -2500/-20

⇒ h = 125 m

∴ Maximum height reached by body = 125 m

$\rule{200}{2}$

$\rule{200}{2}$

Now time taken to reach maximum height :

So using 1st equation of motion :

⇔ v = u + gt

⇒ 0 = 50 + (-10)t

⇒ 0 - 50 = -10t

⇒ - 50 = - 10t

⇒ t = -50/-10

⇒ t = 5 s

∴ Time taken to reach maximum height = 5 s