Question

1. a golf ball is hit at a cliff 150m away with an initial velocity of 70m/s at an angle of 40° to the horizontal.

a) how long will it take for the ball to hit the cliff?
b) at what height did the ball strike the cliff?

1. a cannon is fired horizontally from the top of a 30m cliff into the ocean. the initial velocity of the cannon was 100m/s. how far from the cliff does the cannon - ball hit the water?

Explanation:

Step1: Analyze horizontal - motion for part a

The horizontal component of the initial velocity is $v_{0x}=v_0\cos\theta$, where $v_0 = 70\ m/s$ and $\theta = 40^{\circ}$. The horizontal displacement $x = 150\ m$. In horizontal - motion (no acceleration, $a_x = 0$), the equation is $x=v_{0x}t$. So $t=\frac{x}{v_{0x}}=\frac{x}{v_0\cos\theta}$.
$v_{0x}=70\cos40^{\circ}\approx70\times0.766 = 53.62\ m/s$. Then $t=\frac{150}{53.62}\approx2.8\ s$.

Step2: Analyze vertical - motion for part b

The vertical component of the initial velocity is $v_{0y}=v_0\sin\theta$, where $v_0 = 70\ m/s$ and $\theta = 40^{\circ}$, so $v_{0y}=70\sin40^{\circ}\approx70\times0.643 = 45.01\ m/s$. The equation for vertical displacement is $y = v_{0y}t-\frac{1}{2}gt^{2}$, with $g = 9.8\ m/s^{2}$ and $t\approx2.8\ s$.
$y=45.01\times2.8-\frac{1}{2}\times9.8\times(2.8)^{2}$
$y = 126.028-38.416=87.612\ m$.

Step3: Analyze vertical - motion for part 4

In vertical - motion, the initial vertical velocity $v_{0y}=0\ m/s$, the vertical displacement $y = 30\ m$, and the acceleration $a = g=9.8\ m/s^{2}$. Using the equation $y = v_{0y}t+\frac{1}{2}gt^{2}$, since $v_{0y}=0$, we have $t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2\times30}{9.8}}\approx2.47\ s$.
In horizontal - motion, $v_{0x}=100\ m/s$, and the horizontal displacement $x = v_{0x}t$. So $x = 100\times2.47 = 247\ m$.

Answer:

a) Approximately $2.8\ s$.
b) Approximately $87.6\ m$.

1. $247\ m$.