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A projectile is launched with an initial velocity of v⃗ =(4m/s)x^+(3m/s)y^. What is the velocity of the projectile when it reaches its highest point?

Answer :

abidemiokin

Answer:

4m/s

Explanation

Give the initial velocity of am object as v =(4m/s)xi+(3m/s)yj

From the expression;

Vy = 3m/s

Vx = 4m/s

At the maximum height, the velocity of the body is zero i.e Vy = 0m/s

The velocity of the projectile as it reaches its highest point will be:

v = (4m/s)xi + 0

v = √4²+0²

v = √16

v = 4m/s

Hence he velocity of the projectile when it reaches its highest point is 4m/s

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