Example: The vertical height of a projectile launched from ground level with an initial velocity of \(25m/s\) can be modeled with the equation:
\[ h(t) = 25\sin(\theta)t-4.9t^2 \]Where \(\theta\) is the launch angle and \(t\) is the time in seconds. Suppose the projectile needed to be 25 meters above the ground 2 seconds after launch. What angle should the projectile be launched at?
Solution
To solve this, we set up the equation with the provided parameters: \[ 25 = 25\sin(\theta)\times(2)-4.9(2)^2 \] We need to reorganize and solve this equation for \(\theta\): \[ \solve{ 25 &=& 25\sin(\theta)\times 2-4.9\times 4\\ 25 &=& 50\sin(\theta)-19.6\\ 44.6 &=& 50\sin(\theta)\\ \frac{44.6}{50} &=& \sin(\theta)\\ \sin^{-1}\left(\frac{44.6}{50}\right) &=& \theta\\ 63.13^\circ & \approx &\theta } \] Thus, we find that we need to launch at roughly \(63.13^\circ\) in order to meet the required conditions.