Question

a basketball is tossed upwards with a speed of 5.0 m/s What is the maximum height reached by the basketball from its release point?

Answer 1

Answer:

Maximum height, h = 1.27 meters

Explanation:

It is given that,

Initial speed of the basketball, u = 5 m/s

At maximum height, the final speed of the basketball, v = 0

Let h is the maximum height reached by the basketball from its release point. It can be calculated using third equation of motion as :

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

Here, a = -g

$-u^2=-2gh$

$u^2=2gh$

$h=\dfrac{u^2}{2g}$

$h=\dfrac{(5)^2}{2\times 9.8}$

h = 1.27 meters

So, the maximum height reached by the basketball from its release point is 1.27 meters. Hence, this is the required solution.

Answer 2

Answer:

1.275 m

Explanation:

Let the maximum height reached be h.

Here initial velocity, u = 5 m/s

Final velocity, V = 0

Use third equation of motion

V^2 = u^2 + 2 g h

0 = 25 - 2 × 9.8 × h

h = 25/19.6 = 1.275 m