Question

A stone is thrown with a speed of 10 m s' at an angle of projection 60°. Find its height above the
point of projection when it is at a horizontal distance of 3 m from the thrower? Take g = 10 m 5-27​

Answer 1

here Ux is velocity along x axis
and Uy is velocity along y axis

Answer 2

Given:

• Velocity of projection = 10 m/s.
• Angle of projection = 60°
• Horizontal distance travelled = 3 m

To find:

• Height at that position ?

Calculation:

First of all, latest calculate the time taken by the projectile to cover a horizontal distance of 3 m.

$x = u \cos( \theta) \times t$

$\implies \: 3 = 10 \cos( {60}^{ \circ} ) \times t$

$\implies \: 3 = 5 \times t$

$\implies \: t = \dfrac{3}{5}$

Now, the height attained at that position :

$y = u \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

$\implies y = 10 \sin( {60}^{ \circ} ) \times \dfrac{3}{5} - 5 \times \dfrac{9}{25}$

$\implies y = 3 \sqrt{3} - \dfrac{9}{5}$

$\implies y = 3 \sqrt{3} - 1.8$

$\implies y = 3 .39 \: m$

So,height attained is 3.39 metres .