A NASA test rocket launch is set up so that once the rocket takes off from a landing platform, It will splash down safely in a body of water 5kilometers away. If the launch angle for the experiment is set to an angle of elevation of 30 degrees, what is the minimum speed necessary to make sure the rocket reaches the water? Notes:1 kilometer = 1000meters

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

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answer: 12,000

step-by-step explanation: first you ar e supposed to add. then multiply then subtract