Question

A projectile is launched with a velocity of 13.2 m/s at an angle of 37.0° above the horizontal.
What is the speed of the projectile at its highest point?
a. 7.94 m/s
b. 13.2 m/s
c. 10.5 m/s
d. zero

Answer 1

Answer:

c.

Explanation:

Given that:

The initial speed of the projective v = 13.2 m/s

The angle θ = 37.0°

At the highest point, the particle will comprise only the horizontal component of the speed because the vertical component will be zero.

So,

the horizontal component $v_x = vcos \theta$

$v_x = 13.2 \ m/s (cos 37^0)$

$\mathsf{v_x = 10.5 \ m/s}$