A ball is thrown vertically upward with a speed of $25.0 m / s$ . How high does it rise?

31.9 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

32 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

33 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

33.9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $31.9 m$
Given :
$u = 25 m / s$ (Taking upward direction to be positive)
$v = 0 m / s$
$a = - g = - 9.8 m / s^{2}$
$S = H$
Applying $3 r d$ $e q n$ of motion,
$v^{2} = u^{2} + 2 a S$
$0^{2} = 25^{2} + 2 \times (- g) \times H$
$H = \frac{625}{2 \times 9.8}$
$H = 31.9 m / s$

Suggest Corrections

Similar questions

Q. A ball is thrown vertically upward with a speed of

25.0 m / s

. How long does it take to reach its highest point?

Q. A ball is thrown vertically upward. It has speed of

10 m / s

when it is reached one half of its maximum height. How high does the ball rise? (Taking

g = 10 m / s^{2}

Q. A ball is thrown vertically upwards. It has a speed of

10 m / s e c

, when it has reached one half of its maximum height. How high does the ball rise? Take

g = 10 m / s^{2}

)

Q. A ball is thrown vertically upward with a speed of

25.0 m / s

. How long does the ball take to hit the ground after it reaches its highest point?

A ball is thrown vertically upwards. when the ball reaches one half of its maximum height, its speed becomes 10m/s
(a) How high does the ball rise?
(b) Find the velocity and acceleration 1s after it is thrown.

Join BYJU'S Learning Program

A ball is thrown vertically upward with a speed of 25.0 m/s. How high does it rise?

The correct option is A 31.9mGiven : u=25m/s (Taking upward direction to be positive)v=0m/sa=−g=−9.8 m/s2S=HApplying 3rd eqn of motion,v2=u2+2aS02=252+2×(−g)×HH=6252×9.8H=31.9m/s

A ball is thrown vertically upward with a speed of $25.0 m / s$ . How high does it rise?

The correct option is A $31.9 m$
Given :
$u = 25 m / s$ (Taking upward direction to be positive)
$v = 0 m / s$
$a = - g = - 9.8 m / s^{2}$
$S = H$
Applying $3 r d$ $e q n$ of motion,
$v^{2} = u^{2} + 2 a S$
$0^{2} = 25^{2} + 2 \times (- g) \times H$
$H = \frac{625}{2 \times 9.8}$
$H = 31.9 m / s$