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Chapter 3 Laws Of Motion | Exercise 3E

Question 40

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)

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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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Our top 5% students will be awarded a special scholarship to Lido.

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Question 40

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)

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

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}

Our top 5% students will be awarded a special scholarship to Lido.

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