Question

31. A body is thrown with the velocity u = 12.0 ms' at an angle of a = 45° to the horizon
dropped to the ground at the distance s from the point where it was thrown. From what
height h should stone be thrown in a horizontal direction with the same initial velocity u
for it to fall at the same spot?​

Answer 1

Given:-

$= V_{0} = 12m/s=V$

α = 45

To find :-

h = ?

Solution:-

$= R = S =\sf \frac{12^{2} sin2\alpha }{10}$

$\sf = \frac{144sin90}{10}$

$\sf = 144/10 = 72/5=S=144/10=72/5$

$\sf= u\sqrt{\frac{2h}{g}}$

$S = 12\sqrt{\frac{2h}{g}}$

$H = \frac{144 * 10}{100*2}$

• h = 7.2

Answer 2

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