Question

A ball is thrown vertically upwards with a velocity of 50m/s calculate
a) the maximum height to which it rises
b) the total time it takes to return to the surface of earth. (take g =10m/s)​

Answer 1

Answer:

u=50m/s

g=−10m/s

2

The velocity at the highest point will be zero

Thus, v=0

Height can be calculated as,

=2gs

0−2500=−20s

s=125m

Time can be calculated as,

v=u+gt

0=50+(−10)t

t=5sec

Answer 2

Given :-

• Initial velocity, u = 50m/s

To Find :-

• Maximum height to which it rises.
• Total time taken by the ball to return to the surface of earth.

Solution :-

For a body thrown vertically upward, g is taken negative.

The velocity of the ball at maximum height = 0

Using the third equation of kinematics we can find the maximum height to which the ball rises.

$\boxed{\bf v^2-u^2 = 2gh}$

$\longrightarrow \sf 0^2- 50^2 = 2 \times -10 \times h \\\\\longrightarrow -2500 = -20h \\\\\longrightarrow h = \dfrac{-2500}{-20} \\\\\longrightarrow \bf h = 125$

Maximum height to which the ball rises = 125m

Using the first equation of kinematics we can find the total time taken by the ball to return to the surface of earth.

$\boxed{\bf v = u+gt}$

$\longrightarrow \sf 0 = 50 + -10 \times t \\\\\longrightarrow 0 = 50-10t \\\\\longrightarrow 10t= 50 \\\\\longrightarrow \bf t = 5$

Total time taken by the ball to return to the surface of the earth

                = 2t

                = 10 seconds