Question

A projectile is launched at t=0 with velocity 20m/s at an angle 53°.Find the magnitude of velocity of the projectile at the highest point of the trajectory

Answer 1

Explanation:

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Answer 2

Answer:

The magnitude of velocity of the projectile at the highest point of the trajectory will be 12m/s.

Explanation:

When the particle will reach at highest point then its velocity in the vertical direction will become zero and the net velocity will only be present along horizontal direction. Which is given as,

$u_x=ucos\theta$        (1)

Where,

$u_x$=velocity along horizontal direction

u=the initial velocity with which the body is projected

θ=angle of projection

From the question we have,

u=20m/s

θ=53°

By substituting the required values in equation (1) we get;

$u_x=20\times cos53\textdegree$

$u_x=20\times \frac{3}{5}$

$u_x=4\times 3=12m/s$

Hence, the magnitude of velocity of the projectile at the highest point of the trajectory will be 12m/s.