A body is projected from the ground at an angle of 45 ° with the horizontal. Its velocity after 2 s is 20 m s - 1 . The maximum height reached by the body during its motion is _____ m . (use g = 10 m s - 2 )

Step-by-step solution

After 2   s the velocity is 20   m   s - 1 ⇒ v x 2 + v y 2 = v 2 = 400 ⇒ u 2 cos 2 45 ° + u sin 45 ° - g t 2 = u 2 2 + u 2 - 20 2 = 400 ⇒ u 2 2 + u 2 2 + 400 - 40 2 u = 400 ⇒ u 2 = 40 2 u ⇒ u = 40 2   m   s - 1 Now the maximum height of the projectile is given by, H = u 2 sin 2 θ 2 g ⇒ H = 1600 20 × 2 × 2 = 20   m