Kinematics Problem

A ball is gently dropped from a height of 20 m 20 \text{ m} . If its velocity increases uniformly at the rate of 10 m/s 2 10 \text{ m/s}^{2} , with what velocity will it strike the ground?

Input your answer in m/s \text{ m/s} .


The answer is 20.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

5 solutions

Sravanth C.
May 17, 2015

Here, we have the following values,

H e i g h t = s = 20 m Height=s=20m

A c c e l e r a t i o n = g = 10 m / s 2 Acceleration=g=10m/s^{2}

I n i t i a l v e l o c i t y = 0 m / s Initial \quad velocity=0m/s

F i n a l v e l o c i t y = ? Final \quad velocity=?

Substituting the values in the third equation of motion, i.e. v 2 u 2 = 2 g s v^{2}-u^{2}=2gs

We get, v 2 = 2 × 10 × 20 v^{2}=2×10×20

v = 400 v=\sqrt{400}

Therfore, v = 20 m / s v=\boxed{20m/s}

As every time, excellently written :-) .

Swapnil Das - 6 years ago

Log in to reply

Thanks Swapnil!

Sravanth C. - 6 years ago

if a ball is dropped, doesn't wind resistance mean on a round object such as this its speed remains constant?

Gary Crispin - 5 years, 11 months ago
Ossama Ismail
Apr 6, 2016

K.E = P.E

1 2 m v 2 = m g h \frac{1}{2} m v^2 = m g h

1 2 v 2 = g h = 200 \frac{1}{2} v^2 = g h = 200

v = 20 m / s v = 20 m/s

Nice Solution...

Prokash Shakkhar - 4 years, 6 months ago
Nandik Devbhuti
Aug 4, 2015

distance(s) = 20m initial velocity(u) = 0m final velocity = v acceleration = 10m/s v^2 - u^2 = 2as v^2 = 2 * 10 * 20 v^2 = 400 v = 20. Thus, final velocity is 20 m.

V^2 = 2as

V^2 = 400

V=20 m/s

Shishir Athreyas
May 17, 2015

Very simple just use laws of motion Apply the formulas carefully Here we apply 3rd law of motion V^2-u^2=2ag

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...