Question

for a projectile launched at 54° at an initial height of 1 m and with an initial velocity of 13 m/s, after how many seconds will it have reached its maximum height? 0.55 s 0.85 s 1.07 s 0.41 s

Explanation:

Step1: Find the vertical component of the initial velocity

The vertical component of the initial velocity \(v_{0y}\) is given by \(v_{0y}=v_0\sin\theta\), where \(v_0 = 13\space m/s\) and \(\theta = 54^{\circ}\).

$$v_{0y}=13\sin(54^{\circ})$$

Using a calculator, \(\sin(54^{\circ})\approx0.809\), so \(v_{0y}=13\times0.809 = 10.517\space m/s\)

Step2: Use the kinematic equation for vertical motion at maximum height

At maximum height, the vertical velocity \(v_y = 0\). The kinematic equation \(v_y=v_{0y}-gt\) (where \(g = 9.8\space m/s^{2}\)) can be solved for \(t\).

$$0 = v_{0y}-gt$$

$$t=\frac{v_{0y}}{g}$$

Substitute \(v_{0y}=10.517\space m/s\) and \(g = 9.8\space m/s^{2}\)

$$t=\frac{10.517}{9.8}\approx1.07\space s$$

Answer:

1.07 s