Question

what will be the maximum height obtained by a projectile launched at a 44° angle with an initial height of 0 m and an initial velocity of 11 m/s? 5.27 m 1.26 m 2.98 m 3.64 m

Explanation:

Step1: Find the vertical component of initial velocity

The formula for the vertical component of initial velocity \(v_{0y}\) is \(v_{0y}=v_0\sin\theta\), where \(v_0 = 11\ m/s\) and \(\theta = 44^{\circ}\).
\(v_{0y}=11\times\sin(44^{\circ})\approx11\times0.6947 = 7.6417\ m/s\)

Step2: Use the kinematic equation for vertical motion

The kinematic equation \(v_y^{2}=v_{0y}^{2}-2gh\) is used. At maximum height, \(v_y = 0\). Rearranging for \(h\) gives \(h=\frac{v_{0y}^{2}}{2g}\), where \(g = 9.8\ m/s^{2}\).
Substitute \(v_{0y}=7.6417\ m/s\) into the formula:
\(h=\frac{(7.6417)^{2}}{2\times9.8}=\frac{58.397}{19.6}\approx2.98\ m\)

Answer:

2.98 m