Question

A ball thrown vertically upward returns to the thrower in 20 sec. Find the velocity with which it was thrown and the max. height attained by the ball take g=10m/s2

Answer 1

we know that time of descending = time if ascending
total time taken=20 then time for ascending=10sec
at highest point v=0
g=-10m/s^2(for ascending)

By using first eq of motion
v=u+gt
-u=-gt
u=10×10
u=100m/s

Max height attained
by using third eq of motion
v^2=u^2+2as
u^2=2as
100×100=2×10×s
s=500m

Answer 2

SO, INITIAL VELOCITY IS 100 m/s & MAX HEIGHT IS 500 m.

Given:

• Ball returns back to thrower in 20 sec.

To find:

• Max height
• Initial velocity

Calculation:

We know that the time taken for ascending to max height is equal to time taken to descend back to ground.

So, time for ascent = 20/2 = 10 sec

Now, applying 1st Equation of Kinematics:

$v = u - gt$

$\implies \: 0 = u - (10 \times 10)$

$\implies \: u = 100 \: m {s}^{ - 1}$

Now, max height will be :

${v}^{2} = {u}^{2} - 2gh$

$\implies \: {0}^{2} = {100}^{2} - (2 \times 10 \times h)$

$\implies \: 20h = {100}^{2}$

$\implies \: 20h = 10000$

$\implies \: h = 500 \: m$

So, max height is 500 metres & initial velocity is 100 m/s