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Standard XII

Physics

Horizontal and Vertical Components of Projectile Motion

A ball is thr...

Question

A ball is thrown vertically upwards with an initial velocity of $49 m s^{- 1}$ . Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take $g = 9.8 m s^{- 2}$ )

(i) 122.5 m, (ii) 10 s

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(i) 120.5 m, (ii) 10 s

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(i) 122.5 m, (ii) 20 s

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(i) 120 m, (ii) 10 s

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Solution

The correct option is B (i) 122.5 m, (ii) 10 s
(i)
$v^{2} = u^{2} + 2 a s$
$0 = 49^{2} - 2 (9.8) (s)$
$s = 122.5 m / s$
(ii)
$v = u + a t$
$0 = 49 - 9.8 t$
$t = 5 s$
total time $= 10 s$

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A ball is thrown vertically upwards with an initial velocity of 49ms−1. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g=9.8ms−2)

The correct option is B (i) 122.5 m, (ii) 10 s(i) v2=u2+2as0=492−2(9.8)(s)s=122.5 m/s(ii)v=u+at0=49−9.8tt=5 stotal time=10 s

A ball is thrown vertically upwards with an initial velocity of $49 m s^{- 1}$ . Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take $g = 9.8 m s^{- 2}$ )

The correct option is B (i) 122.5 m, (ii) 10 s
(i)
$v^{2} = u^{2} + 2 a s$
$0 = 49^{2} - 2 (9.8) (s)$
$s = 122.5 m / s$
(ii)
$v = u + a t$
$0 = 49 - 9.8 t$
$t = 5 s$
total time $= 10 s$