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Selina Solutions Class 9 Physics Solutions for Exercise 3E in Chapter 3 - Chapter 3 Laws Of Motion

Question 47 Exercise 3E

A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6m/s. the ball reaches the ground after 5s. Calculate: (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g=9.8m/s^2

Answer:

Given:

Initial velocity=19.6m/s, t=5s;

Velocity at the highest point is zero

(i) To calculate the height of the tower

Assume‘d’ to be the height of the tower and ‘h’ to be the distance from the top of the tower to the maximum height as shown in the figure.

Selina Solutions CONCISE Physics - Class 9 ICSE chapter Chapter 3 Laws Of Motion Question 47 Solution image

We know from the equation of motion,

\begin{array}{l} v^{2}-u^{2}=2 g h \\ \quad \Rightarrow \quad 0-(19.6)^{2}=2(9.8)(h) \\ \quad \Rightarrow \quad h=19.6 \mathrm{m} \end{array}

(ii)

Let t1 be the time taken by the ball to reach the greatest point from the top of the tower

To calculate the time for which the ball remains in the air;

We know the from the equation of motion,

v=u-gt

0 = 19.6 – (9.8)(t1)

t1 = 2s

Assume motion for (h+d) part;

Time taken for the ball to reach from the highest point of the tower to the ground is

t-t1 = 5-2 = 3s

From the equation of motion,

0 = 19.6 – (9.8)(t1)

t1 = 2s

Assume motion for (h+d) part;

Time taken for the ball to reach from the highest point of the tower to the ground is

t-t1 = 5-2 = 3s

From the equation of motion,

s=u t+1 / 2 g t^{2}

‘s’ here is the distance from the top of the tower to the highest point, i.e., h+d \begin{array}{l} \Rightarrow \mathrm{h}+\mathrm{d}=0+1 / 2(9.8)(3)^{2} \\ \Rightarrow \mathrm{d}+19.6=44.1 \mathrm{m} \\ \Rightarrow \mathrm{d}=24.5 \mathrm{m} \end{array}

The height of the tower is 24.5m

(iii) Assume ‘v’ be the velocity of the ball when it strikes the ground

We know from the relation;

v = u +gt

v = 0 + (9.8)(3)

= 29.4m/s

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