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…

Question

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))

MARKS Content Team · Accepted Answer

Correct answer: 20. After 2 s the velocity is 20 m s^(-1) ⇒vx2+vy2=v^2=400 ⇒u^2cos^245°+usin45°-gt^2=u^22+u2-20^2=400 ⇒u^22+u^22+400-(40√2u=400 ⇒u^2=40√2u⇒u=40√2 m s^(-1)Now the maximum height of the projectile is given by,H=u^2sin^2θ2g⇒H=1600)/(20×2×2)=20 m

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 $g = 10 {m s}^{- 2}$ )

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Relevant Questions

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)

Solution

Relevant Questions

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 $g = 10 {m s}^{- 2}$ )