Question

A ball is thrown vertically upwards with an initial velocity of 54 km/h. Calculate the maximum

height attained by it (take g = 9.8 m/s2

).​

Answer 1

Provided that:

• Initial velocity = 54 kmph
• Final velocity = 0 mps
• g = -9.8 mps sq.

Don't be confused!

• Final velocity cames as zero because the ball is thrown upwards and it will stop at the highest point.

• We take g as negative because the ball is thrown upwards that is against the gravity.

To calculate:

• Maximum height attained

Solution:

• Maximum height attained = 11.47 m

Using concepts:

• Formula to convert kmph-mps.
• Third equation of motion.

Using formulas:

• Formula to convert kmph-mps:

• ${\small{\underline{\boxed{\pmb{\sf{1 \: kmph \: = \dfrac{5}{18}}}}}}}$

• Third equation of motion:

• ${\small{\underline{\boxed{\pmb{\sf{v^2 \: - u^2 \: = 2as}}}}}}$

Where, v denotes final velocity, u denotes initial velocity, s denotes displacement or distance or height and a denotes acceleration.

Required solution:

~ Firstly let us convert initial velocity into metre per second!

$\begin{gathered}:\implies \sf 54 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{54} \times \dfrac{5}{\cancel{{18}}} \: (Cancelling) \\ \\ :\implies \sf 3 \times 5 \\ \\ :\implies \sf 15 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}\end{gathered}$

• Initial velocity = 15 mps!

~ Now by using third equation of motion let us find out the maximum height!

$:\implies \sf v^2 \: - u^2 \: = 2as \\ \\ :\implies \sf (0)^{2} - (15)^{2} \: = 2(-9.8)(s) \\ \\ :\implies \sf 0 - 225 = -19.6s \\ \\ :\implies \sf -225 = -19.6s \\ \\ :\implies \sf 225 = 19.6s \\ \\ :\implies \sf \dfrac{225}{19.6} \: = s \\ \\ :\implies \sf 11.47 \: m \: = s \\ \\ :\implies \sf Height \: = 11.47 \: m$

Answer 2

Given Initial velocity of ball, u=49 m/s
Let the maximum height reached and time taken to reach that height be H and t respectively.
Assumption: g=9.8 m/s
2
holds true (maximum height reached is small compared to the radius of earth)

Velocity of the ball at maximum height is zero, v=0

v
2
−u
2
=2aH
0−(49)
2
=2×(−9.8)×H
⟹H=122.5 m

v=u+at
0=49−9.8t
⟹t=5 s
∴ Total time taken by ball to return to the surface, T=2t=10 s