Answer:
237.87 m/s
Step-by-step explanation:
This is a projectile question.
We use the equation for the range of a projectile since we are given the horizontal distance and angle of projection.
So, R = U²sin2θ/g where R = range, U = initial speed of rocket, θ = projection angle and g = acceleration due to gravity = 9.8 m/s².
Since it is given that the water body is 5 km away, R = 5 km = 5000 m and θ = 30°
So, making U subject of the formula, we have that
U = √(gR/sin2θ)
So, substituting the values of the variables, we have
U = √[9.8 m/s² × 5000 m/sin(2 × 30°)]
U = √[49000 m²/s²/sin60°]
U = √[49000 m²/s²/0.8660]
U = √[56,580.33 m²/s²]
U = 237.87 m/s
So the minimum speed of the rocket necessary to make sure the rocket reaches the water surface is U = 237.87 m/s
