Question

A ball is thrown vertically upwards from ground when it is 3/4 of its maximum height its speed is 20 m/s .The speed of throw is​

Answer 1

Answer:

40 m/sec

Explanation:

refer to attachment

Answer 2

Given:

When the ball is at 3/4 of its maximum height then, speed of ball=20 m/s

To find:

Speed of the throw

Solution:

Let initial speed of ball=u

Maximum height reached by ball=h

Speed at maximum height=0

Now,

$v^2-u^2=2gh$...(1)

Using the formula

$0-u^2=-2gh$

Because we are going against gravity therefore, we are taking g is negative

$-u^2=-2gh$

$u^2=2gh$

Height=3/4h

Speed,v=20 m/s

Again, using equation (1)

$(20)^2-u^2=-2g\times \frac{3}{4}h$

$400-(2gh)=-\frac{3}{2}gh$

$400-2gh=-\frac{3}{2}gh$

$400=-\frac{3}{2}gh+2gh$

$400=\frac{-3+4}{2}gh$

Where $g=10ms^{-2}$

$400=\frac{1}{2}\times 10 h$

$h=\frac{400\times 2}{ 10}$

h=80 m

Substitute the value

$u^2=2\times 10\times 80=1600$

$u=\sqrt{1600}=40 m/s$

Hence, speed of throw=40 m/s