Question

An object is launched at a speed of 15 m/s. If it rises to a height of 1 m, at what angle was it launched?

Answer 1

Answer:

Angle at which object is launched is 17.15°

Explanation:

We have given initial velocity at which object is launched u = 15 m/sec

It rises to a height of 1 m

So height h = 1 m

We have to find the angle of projection $\Theta$

Acceleration due to gravity $g=9.8m/sec^2$

We know that maximum height is given by $h=\frac{u^2sin^2\Theta }{2g}$

So $1=\frac{15^2\times\ sin^2\Theta }{2\times 9.8}$

$sin^2\Theta =0.0871$

$sin\Theta =0.295$

$\Theta =sin^{-1}0.295=17.15^{\circ}$

So the angle at which object is launched is 17.15°