Question

What is the height at the peak of the projectiles trajectory if the magnitude vo equals 42.2 meters per second, and the angle ofΘis 41.8 degrees?

Answer 1

Take into account that the vertical final velocity of the projectile is given by:

$v^2_y=v^2_{oy}-2gy$

Now, consider that at the peak of the trajectory, the vertical final velocity is zero. Moreover:

$v^2_{oy}=v^2_o\sin ^2\theta$

Then, by solving for y into the equation for vy^2, you obtain:

$\begin{gathered} o=v^2_o\sin ^2\theta-2gy \\ y=\frac{v^2_o\sin \theta}{2g} \end{gathered}$

where g = 9.8m/s^2 is the gravitational acceleration constant.

By using the given values for the angle and vo you obtain for the height of the peak y:

$y=\frac{(42.2\frac{m}{s})(\sin 41.8)^2}{2(9.8\frac{m}{s^2})}\approx40.36m$

Hence, the height of the peak of the projectil trajectory is approximately 40.36 m