Question

1. a projectile is fired from the edge of a cliff 95 m high with an initial speed of 50 m/s at an angle of 37° above the horizontal. calculate the following:

a. the total time in the air.
b. the maximum horizontal distance covered by the projectile.
c. the maximum height reached by the projectile relative to the ground.

Explanation:

Step1: Analyze vertical motion

The vertical component of initial velocity \(v_{0y}=v_0\sin\theta\), where \(v_0 = 50\space m/s\) and \(\theta = 37^{\circ}\), so \(v_{0y}=50\sin37^{\circ}=50\times0.6 = 30\space m/s\). The vertical displacement \(y=- 95\space m\) (taking downwards as negative), and the acceleration \(a=-g=- 9.8\space m/s^{2}\). Using the equation \(y = v_{0y}t+\frac{1}{2}at^{2}\), we get \(-95 = 30t-4.9t^{2}\), or \(4.9t^{2}-30t - 95=0\).

Step2: Solve quadratic equation for time

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 4.9\), \(b=-30\), \(c=-95\)), the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substituting the values: \(t=\frac{30\pm\sqrt{(-30)^{2}-4\times4.9\times(-95)}}{2\times4.9}=\frac{30\pm\sqrt{900 + 1862}}{9.8}=\frac{30\pm\sqrt{2762}}{9.8}=\frac{30\pm52.55}{9.8}\). We take the positive root \(t=\frac{30 + 52.55}{9.8}\approx8.42\space s\).

Step3: Calculate horizontal distance

The horizontal component of velocity \(v_{0x}=v_0\cos\theta=50\cos37^{\circ}=50\times0.8 = 40\space m/s\). Using \(x = v_{0x}t\), with \(t = 8.42\space s\), we get \(x=40\times8.42 = 336.8\space m\).

Step4: Find maximum height

At maximum height, \(v_y = 0\). Using \(v_y^{2}-v_{0y}^{2}=2a\Delta y\), \(\Delta y=\frac{v_y^{2}-v_{0y}^{2}}{2a}=\frac{0 - 30^{2}}{2\times(-9.8)}=\frac{-900}{-19.6}\approx45.92\space m\). The height relative to the ground \(H=95 + 45.92=140.92\space m\).

Answer:

a. The total time in the air is approximately \(t = 8.42\space s\).
b. The maximum horizontal distance is \(x = 336.8\space m\).
c. The maximum height relative to the ground is \(H = 140.92\space m\).