Answer to Question #158619 in Physics for Shelly

Question

The highest barrier that a projectile can clear is 10.0 m, when the projectile is launched at an angle of 29.0 ° above the horizontal. What is the projectile's launch speed?

Answer 1

Let's first find the $y$-component of the initial velocity of the projectile:

$$v_y^2=v_{0y}^2-2gy,$$ $$0=v_{0y}^2-2gy,$$ $$v_{0y}=\sqrt{2gy}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot10.0\ m}=14\ \dfrac{m}{s}.$$

Finally, we can find the projectile's launch speed from the trigonometry:

$$v_0=\dfrac{v_{0y}}{sin\theta}=\dfrac{14\ \dfrac{m}{s}}{sin29.0^{\circ}}=28.87\ \dfrac{m}{s}.$$