Question

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal. At its highest point in this trajectory, the velocityof the projectile would be what?

Answer 1

v₂ = 176.24 m/s

Step-by-step explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

`H_(max)= (v^2 sin^2(\theta))/(g)`

now,

`H_(max)= (v_1^2 sin^2(\theta_1))/(g)`

`H_(max)= (v_2^2 sin^2(\theta_2))/(g)`

now, equating both the equations

`(v_2^2)/(v_1^2)=(sin^2(\theta_1))/(sin^2(\theta_2))`

`(v_2^2)/(150^2)=(sin^2(45^0))/(sin^2(37^0))`

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s