Question

A projectile is launched at an angle of 45 degrees with a velocity of 250 m/s. If air resistance is neglected, the magnitude of the horizontal velocity of the projectile at the time it reaches maximum altitude is equal to

Answer 1

Answer:

The horizontal velocity of the projectile at the time it reaches maximum altitude approximately equal to 131.331 meters per second.

Explanation:

From Mechanical Physics we understand that projectile motion is the combination of a horizontal motion at constant velocity and a vertical uniform accelerated motion due to gravity. Then, the horizontal velocity of the projectile ($v_{x}$), measured in meters per second,  remains constant during motion. That is:

$v_{x} = v_{o}\cdot \cos \theta$ (Eq. 1)

Where:

$v_{o}$ - Initial velocity of the projectile, measured in meters per second.

$\theta$ - Launch angle above the horizontal, measured in sexagesimal degrees.

If we know that $v_{o} = 250\,\frac{m}{s}$ and $\theta = 45^{\circ}$, then horizontal velocity of the projectile at the time it reaches maximum altitude is:

$v_{x} = \left(250\,\frac{m}{s} \right)\cdot \cos 45^{\circ}$

$v_{x} \approx 131.331\,\frac{m}{s}$

The horizontal velocity of the projectile at the time it reaches maximum altitude approximately equal to 131.331 meters per second.