Question

Q8/180 A projectile is launched at time t = 0
with velocity 20 m/s at an angle 53° with the
horizontal. The magnitude of velocity of the
projectile at the highest point of the trajectory is​

Answer 1

Answer: 18.36 (approximate)

Explanation: magnitude of velocity at the highest point in a projectile is

u•cos•Φ

= 20m/s x cos 53

= -18.36

Since magnitude is asked, the answer is 18.36

Answer 2

Answer:

The magnitude of velocity of the projectile at the highest point of the trajectory is​ 12m/s.

Explanation:

The magnitude of velocity of the projectile at the highest point of the trajectory will only be along the horizontal direction which is given as,

$u_x=ucos\theta$         (1)

Where,

$u_x$=velocity at the highest point

u=initial velocity with which the body is thrown

θ=angle of projection

From the question we have,

u=20m/s

θ=53°

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

$u_x=20\times cos53\textdegree$

$u_x=20\times \frac{3}{5}$                 ($cos53\textdegree=\frac{3}{5}$)

$u_x=12m/s$

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