Question

A cannon tilted upward at 30° fires a cannonball with a speed of 100 m/s. At that instant, what is the component of the cannonball’s velocity parallel to the ground?

Answer 1

Answer:

the cannonball’s velocity parallel to the ground is 86.6m/S

Explanation:

Hello! To solve this problem remember that in a parabolic movement the horizontal component X of the velocity of the cannonball is constant while the vertical one varies with constant acceleration.

For this case we must draw the velocity triangle and find the component in X(see atached image).

V= Initial velocity=100M/S

$cos30=\frac{Vx}{V}$

V= Initial velocity=100M/S

Vx=cannonball’s velocity parallel to the ground

Solving for Vx

Vx=Vcos30

Vx=(100m/S)(cos30)=86.6m/s

the cannonball’s velocity parallel to the ground is 86.6m/S

Answer 2

The component of the cannonball’s velocity parallel to the ground is 50m/s.

The cannonball will follow projectile motion. It is given that the initial velocity of the cannonball is v = 100m/s and it makes an angle of θ = 30° from the horizontal.

Now the component of velocity parallel to the ground is the horizontal component of velocity given by:

$v_x=vcos\theta$  represented in the image attached.

so,

$v_x=100cos60\\\\v_x=100*\frac{1}{2}m/s\\ \\v_x=50m/s$is the horizontal velocity.

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