A projectile with speed 50   m   s - 1 is thrown at an angle of 60 ° with the horizontal. The maximum height that can be reached is, (acceleration due to gravity = 10   m   s - 2 )

Question

A projectile with speed 50 &#160; m &#160; s - 1 is thrown at an angle of 60 &#176; with the horizontal. The maximum height that can be reached is, (acceleration due to gravity = 10 &#160; m &#160; s - 2 )

Aicharya · Accepted Answer

Maximum height of the projectile is given by, $H = \frac{u^{2} {(\sin θ)}^{2}}{2 g}$ .
Given:
$u = 50 m s^{- 1}$ & $θ = 60 °$ .
$H = \frac{{(50)}^{2} {(\sin 60 °)}^{2}}{2 \times 10} = \frac{2500 \times {(\frac{\sqrt{3}}{2})}^{2}}{20} = 93.75 m$

A projectile with speed 50   m   s - 1 is thrown at an angle of 60 ° with the horizontal. The…

Solution

A projectile with speed 50 &#160; m &#160; s - 1 is thrown at an angle of 60 &#176; with the horizontal. The…

Solution

A projectile with speed 50 m s - 1 is thrown at an angle of 60 ° with the horizontal. The…