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Standard IX

Physics

Derivation of Velocity-Time Relation by Graphical Method

A body thrown...

Question

A body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an interval of 10s. The height h is $(g = 10 {m / s}^{2})$

22 m

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10.2 m

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11.2 m

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15 m

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Solution

The correct option is B 10.2 m
Initial velocity, $V = 52 m / s$
Time taken, $t = 10 s e c$
$\Rightarrow S = u t + \frac{1}{2} a t^{2}$
$= 52 (10) - \frac{1}{2} (10) (100)$
$= 20 m$
As, $S = 2 h$
$\Rightarrow h = 10 m$
Hence, the answer is $10 m .$

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A body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an interval of 10s. The height h is (g=10m/s2)

The correct option is B 10.2 mInitial velocity, V=52m/sTime taken, t=10sec⇒S=ut+12at2 =52(10)−12(10)(100) =20mAs, S=2h⇒h=10mHence, the answer is 10m.

A body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an interval of 10s. The height h is $(g = 10 {m / s}^{2})$

The correct option is B 10.2 m
Initial velocity, $V = 52 m / s$
Time taken, $t = 10 s e c$
$\Rightarrow S = u t + \frac{1}{2} a t^{2}$
$= 52 (10) - \frac{1}{2} (10) (100)$
$= 20 m$
As, $S = 2 h$
$\Rightarrow h = 10 m$
Hence, the answer is $10 m .$