Question

A stone thrown with the velocity v0 = 12 m/s at an

angle = 45° to the horizontal, dropped to the

ground at a distance s from the point where it was

thrown . From what height (in metre) h should the

stone be thrown in a horizontal direction with the

same initial velocity v0

for it to fall at the same

spot?​

Answer 1

Given:-

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

α = 45

To find :-

h = ?

Solution:-

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

$= S = \frac{144 * sin90}{10}$

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

$As$

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

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

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

h = 7.2

Thank you....

Answer 2

Step-by-step explanation:

aga solve kar loo please