Question

A ball is thrown with velocity of 10 m/s upwards. If the ball is caught 1 m above its initial position, what is the speed of the ball when it is caught?

Answer 1

Answer:

v = 8.96 m/s

Explanation:

Initial speed of the ball, u = 10 m/s

It caught 1 meter above its initial position.

Acceleration due to gravity, $g=-9.8\ m/s^2$

We need to find the final speed of the ball when it is caught. Let is equal to v. To find the value of v, use third equation of motion as :

$v^2-u^2=2as$

$v^2=2as+u^2$

$v^2=2(-9.8)\times 1+(10)^2$

v = 8.96 m/s

So, the speed of the ball when it is caught is 8.96 m/s. Hence, this is the required solution.

Answer 2

Answer:

8.96 m/s, upward direction

Explanation:

Given that, the initial velocity of the ball is,

$u=10m/s$

And the acceleration in the downward direction is positive but in this situation the acceleration will be negative so,

$a=9.8\frac{m}{s^{2} }$

And according to question vertical displacement is,

$s=1m$

Now suppose v be the final velocity of the ball.

Applying third equation of motion,

$v^{2}=u^{2}+2as$

Here, u is the initial velocity, a is the acceleration, s is the displacement.

Substitute all the variables.

$v=\sqrt{10^{2}+2(-9.8)\times 1 } \\v=\sqrt{80.4}\\ v=8.96\frac{m}{s}$

Therefore, the speed of ball when it is caught is 8.96 m/s in the upward direction.