Question

A stone is thrown horizontally from the top of a tower 50ft. high at 70ft/sec. Find the velocity of the stone as it strikes the ground.

A horizontal stream of water leaves a nozzle 1.5m above the ground and strikes the ground 4m away. Find the initial speed of the water and the speed as strikes the ground.

Answer 1

Question 23382

1. Let $oy$ axis has the same direction as $\vec{g}$. Then, $y$ component of velocity $v_y(t) = v_0 + gt$, where $v_0 = 70\frac{ft}{s} = 70 \cdot 0.3\frac{m}{s} = 21\frac{m}{s}$ and $g = 10\frac{m}{s}$. Integrating formula for $y$ component of velocity, obtain: $y(t) = -h + v_0t + \frac{gt^2}{2}$, $h = 50ft = 15m$. For moment when stone stops: $0 = -h + v_0t + \frac{gt^2}{2}$; $-15 + 21t + 5t^2 = 0 \Rightarrow t \approx 0.62s$. Hence, velocity at this moment is $v_y = 21\frac{m}{s} + 10 \cdot 0.62 = 27.2\frac{m}{s}$.

2. Let $oy$ axis be vertically up.

Then, equations of motion are: $v_x(t) = v; v_y(t) = -gt$, $x(t) = vt; y(t) = h - \frac{gt^2}{2}$.

Let $t_1$ be the moment when water strikes the ground. For this moment, $h = \frac{gt_1^2}{2} \Rightarrow t_1 = \sqrt{2\frac{h}{g}}$, when for x-component of motion $4 = vt_1 \Rightarrow v = \frac{4}{t_1} = \sqrt{8\frac{g}{h}} \approx 7.3\frac{m}{s}$.

Speed, when water strikes the ground is $v = \sqrt{v_x^2 + v_y^2} \mid_{t=t_1} = \sqrt{v^2 + 2gh} = 9.13\frac{m}{s}$. Initial speed is $v = 7.3\frac{m}{s}$.