Question

Kaplit and Mukesh are playing with two different balls of masses m and 2m respectively. If Kaplit throws his ball vertically up and Mukesh at an angle both of them stay in our view for the same period. The height attained by the two balls are in the ratio-

Answer 1

Hint: Kaplit throws the ball straight up, it will go achieve some height, stops there and then retraces its trajectory back to the earth. On the other hand, Mukesh throws the ball at an angle, so this will be the case of projectiles.

Complete step by step answer:
For kaplit:
At its highest point, velocity will be zero. Let u be the velocity of the ball.
\[\begin{align}
  & {{v}^{2}}-{{u}^{2}}=2as \\
 & \Rightarrow 0-{{u}^{2}}=-2\times g\times s \\
 & \Rightarrow s=\dfrac{{{u}^{2}}}{2g} \\
\end{align}\]
\[s=\dfrac{{{u}^{2}}}{2g}\]--(1)
For Mukesh: let the initial velocity of the ball be v.
Let Angle made with the horizontal is \[\psi \]. Then the formula for maximum height is given as \[H=\dfrac{{{v}^{2}}{{\sin }^{2}}\psi }{2g}\]--(2)
Now we need to find the ratio of the height attained by the two balls.
So dividing eq (1) by (2), we get,
\[\dfrac{s}{H}=\dfrac{\dfrac{{{u}^{2}}}{2g}}{\dfrac{{{v}^{2}}{{\sin }^{2}}\psi }{2g}}=\dfrac{{{u}^{2}}}{{{v}^{2}}{{\sin }^{2}}\psi }\]
Thus, the ratio comes out to be \[\dfrac{s}{H}=\dfrac{{{u}^{2}}}{{{v}^{2}}{{\sin }^{2}}\psi }\]

Additional information-: while doing problems involving Newton’s equations of motion, we have to keep in mind the sign conventions. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is deaccelerating and its acceleration is negative. Also, all the units to be used must be in SI units to avoid any mistake. This has to be particularly noted that time can never be negative and if in our answer time is coming in negative then it means we have committed a mistake.

Note:The ball is thrown making an angle with the horizontal. The path covered by the body is called projectile. Always in the formula the angle used is made with the vertical and if in the question it is given that angle is made with the vertical than we have to just subtract the given angle from \[{{90}^{0}}\]