Question

A ball is thrown with a speed of 20 m s–1 in a direction 30° above the horizontal. Calculate (i) the
maximum height attained by the ball and (iii) the time taken by the ball to return to the same level
(Take g = 10 m s–2).

Answer 1

Answer on Question #53587, Physics Mechanics Kinematics Dynamics

A ball is thrown with a speed of $20\,\mathrm{m\,s^{-1}}$ in a direction $30{}^{\circ}$ above the horizontal. Calculate

(i) the maximum height attained by the ball and

(iii) the time taken by the ball to return to the same level

(Take $g = 10\,\mathrm{m\,s^{-2}}$).

Solution

Fig. 1

Maximum lift height is found from the law of conservation of energy. The law of conservation of energy transfer from the position O in A is given by Eq(1).

$$\frac{M v_0^2}{2} = M g H_{\max} + \frac{M v_{0x}^2}{2}$$

where $v_{0x} = v_0 \cos \alpha$; $M$ is the mass of the ball.

Then

$$H_{\max} = \frac{v_0^2 - v_{0x}^2}{2g} = \frac{v_0^2 - v_0^2 \cos^2 \alpha}{2g} = \frac{v_0^2 \sin^2 \alpha}{2g} = \frac{20^2 \sin^2 30^0}{2 \cdot 10} = 5\,\mathrm{m}$$

Time of the ball motion from point O to point A is the same as from point A to point B (the symmetry of the trajectory).

Time of the ball motion from point O to point A (see Fig.1):

$$v_{0y} - g t_0 = 0 \Rightarrow t_0 = v_{0y} / g = v_0 \sin \alpha / g$$

The time taken by the ball to return to the same level is given by Eq.(3)

$$t = 2 t _ {0} = \frac {2 v _ {0} \sin \alpha}{g} = \frac {2 \cdot 2 0 \cdot \sin 3 0 ^ {0}}{1 0} = 2 s$$

Answer: $H_{\max} = \frac{v_0^2\sin^2\alpha}{2g} = 5m; t = \frac{2v_0\sin\alpha}{g} = 2s.$

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