Question

A stone is thrown with a speed of 49 m/& at an angle of 45 upward with the earth horizontal surface. Calculate the horizontal range of the stone.

Answer 1

Explanation:

We have,,

Initial speed of the stone,u = 49 m/s

And angle of projection,theta(∅) = 45°

Therefore, Horizontal range R of the stone is given by,

$\boxed{\sf{R = \frac{u^{2}\sin(2\Theta)}{g}}}$

$\implies{\sf{R = \frac{(49)^{2}\sin(2\times{45°})}{9.8}}}$

$\implies{\sf{R = \frac{2401\times{\sin(90°)}}{9.8}}}$

$\implies{\sf{R = \frac{{2401}\times{1}}{9.8}}}$

$\implies{\sf{R = \frac{{49}\times{10}}{2}}}$

$\implies{\sf{R = \frac{{490}}{2}}}$

$\implies{\boxed{\sf{R = 245 m}}}$

Hence, range of the stone is 245 m ✓✓✓

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