- Chapter 1 – Measurements and Experimentation
- Chapter 2 Motion In One Dimension
- Chapter 3 Laws Of Motion
- Chapter 4 Pressure In Fluids And Atmospheric Pressure
- Chapter 5 Upthrust In Fluids Archimedes Principle And Floatation
- Chapter 6 Heat And Energy
- Chapter 7 Reflection Of Light
- Chapter 8 Propogation Of Sound Waves
- Chapter 9 Current Electricity
- Chapter 10 Magnetism
Selina Solutions Class 9 Physics Solutions for Exercise 3E in Chapter 3 - Chapter 3 Laws Of Motion
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A ball is thrown vertically upwards. It returns 6s later. Calculate: (i) the greatest height reached by the ball, and (ii) the initial velocity of the ball. (Take g=10m/s^2)
Given: time=6s, g=10m/s^2
(i) To calculate the greatest height reached by the ball.
Assume ‘h’ to be the greatest height reached. Ascent
Time for the rise of the ball = 6/2 =3s
Initial velocity for the rise is zero
As per the equation of motion;
\begin{array}{l} \mathrm{h}=\mathrm{ut}+1 / 2 \mathrm{gt}^{2} \\ =0(3)+1 / 2(10)(3)^{2} \\ =0+45 \\\Rightarrow 45 \mathrm{m} \end{array}
(ii) To calculate the initial velocity of the ball
Assume u1 to be the initial velocity
We know from the equation of motion
\begin{array}{l} \mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{gh} \\ \Rightarrow \mathrm{v}^{2}-0=2(10)(45) \\ \quad \Rightarrow \mathrm{v}=30 \mathrm{m} / \mathrm{s} \end{array}
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