Question

daily schedule
10/30(thur) ; for: q2.1 tes
phy: q2.1 test
①
launched at 45° to horizontal, hits a peak of 60m.
① what is ( t_{\text {total }} ) ?
② what is ( d x )

Explanation:

Step1: Find the initial vertical velocity

At the peak of the projectile motion, the vertical velocity \(v_y = 0\). Using the kinematic equation \(v_y^2=v_{0y}^2 - 2gy\) (where \(v_y = 0\), \(y = 60m\) and \(g = 9.8m/s^2\)).

$$0 = v_{0y}^2-2\times9.8\times60$$

$$v_{0y}^2=2\times9.8\times60$$

$$v_{0y}=\sqrt{2\times9.8\times60}\approx34.3m/s$$

Since the launch angle \(\theta = 45^{\circ}\), and \(v_{0y}=v_0\sin\theta\), then \(v_0=\frac{v_{0y}}{\sin\theta}=\frac{34.3}{\sin45^{\circ}}\approx48.5m/s\)

Step2: Find the time to reach the peak

Using the kinematic equation \(v_y = v_{0y}-gt\). At the peak \(v_y = 0\), so \(t=\frac{v_{0y}}{g}=\frac{34.3}{9.8}= 3.5s\)
The total time of flight \(t_{total}=2t\) (symmetry of projectile motion for same - height launch and landing)

$$t_{total}=2\times3.5 = 7s$$

Step3: Find the horizontal velocity

\(v_{0x}=v_0\cos\theta\), since \(v_0\approx48.5m/s\) and \(\theta = 45^{\circ}\), \(v_{0x}=48.5\cos45^{\circ}\approx34.3m/s\)

Step4: Find the horizontal distance \(dx\)

Using the equation \(dx = v_{0x}t_{total}\)

$$dx=34.3\times7 = 240.1m$$

Answer:

a. \(t_{total}=7s\)
b. \(dx\approx240m\)