A rocket is fired at a 45° angle, what is the direction of the horizontal velocity vector at the peak height? Vertically downwards
Horizontally to the left
Upwards at a 45-degree angle
Downwards at 45-degree angle
Zero, it’s not moving in this direction at the peak

A rocket is fired at a 45° angle, what is the direction of the vertical velocity vector at the peak height?

Vertically downwards
Horizontally to the left
Upwards at a 45-degree angle
Downwards at 45-degree angle
Zero, it’s not moving in this direction at the peak

A rocket is fired at a 45° angle, what is the direction of the acceleration vector at the peak height?

Vertically downwards
Horizontally to the left
Upwards at a 45-degree angle
Downwards at 45-degree angle
Zero, it’s not accelerating in this direction at the peak

B: Horizontally to the left

Explanation:

Horizontal velocity is always constant throughout the entire trajectory of the rocket and acts in the horizontal direction in which the rocket was launched. This is because gravity only acts in the downwards vertical direction.

So in order words at peak height, horizontal velocity is in the horizontal direction in which the rocket was launched.

So if it was to the left, then direction is left but if right, then direction is right.

Looking at the options, the most appropriate will be:

Horizontally to the left

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000

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