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Question

# When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity, g = 9.8 m/s2​)

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Solution

## Distance travelled by the ball = 19.6 m Acceleration due to gravity = ​g = 9.8 m/s2​ Let the initial velocity of the body = u Final velocity of the ball = 0 m/s Time taken by the ball to reach to the highest point = t Using the third equation of motion, ${v}^{2}={u}^{2}-2gs$ $⇒0={u}^{2}-2g×19.6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{u}^{2}=2×9.8×19.6=\left(19.6{\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒u=19.6\mathrm{m}/\mathrm{s}$ Now, using the first equation of motion for the ball thrown upward, v = u - gt $\therefore 0=u-gt\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒t=\frac{u}{g}=\frac{19.6\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^{2}}=2\mathrm{s}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒t=2\mathrm{s}$ Time taken by the ball to reach the highest point = 2 s And the initial velocity of the ball, u = 19.6 m/s

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