Question

a man throws a stone to hit a box located 75 m from him.the initial point and the box are in same plane.the initial velocity of the stone is 35 m/s.what is the angle of throwing the stone to hit the box?at exact midpoint of the throwing point and the box,a buckle of tree is at a height of 3.5 m from the throwing point.will the stone go up of the buckle or down of the buckle?

Answer 1

Answer on Question#49931 – Physics – Mechanics | Kinematics | Dynamics

$$\theta = \frac {1}{2} a \sin \left(\frac {7 5 g}{V ^ {2}}\right) \approx 1 8. 8 8 {}^ {\circ}$$

Stone will go up of the buckle.

Solution

Equations of motion:

$$m \frac {d ^ {2} x}{d t} = 0$$ $$m \frac {d ^ {2} y}{d t ^ {2}} = - m g$$

Initial conditions:

$$x = 0; \quad V _ {x} (0) = V \cos (\theta); \quad y (0) = 0; \quad V _ {y} (0) = V \sin (\theta);$$

Explicit solutions of the equations above:

$$x = C _ {1} + C _ {2} t$$ $$y = - \frac {g t ^ {2}}{2} + C _ {3} t + C _ {4}$$

Use initial conditions:

$$x (0) = C _ {1} + C _ {2} (0) = C _ {1} = 0$$ $$V _ {x} (0) = \frac {d x}{d t} (0) = C _ {2} = V \cos (\theta)$$ $$y(0) = - \frac{g(0)^2}{2} + C_3(0) + C_4 = C_4 = 0$$ $$V_y(0) = \frac{dy}{dt}(0) = -g(0) + C_3 = C_3 = V \sin(\theta)$$

Thus, we have:

$$x(t) = V \cos(\theta) t$$ $$y(t) = V \sin(\theta) t - \frac{g t^2}{2}$$

We know that at some time $t_1$:

$$x(t_1) = V \cos(\theta) t_1 = 75$$ $$y(t_1) = V \sin(\theta) t_1 - \frac{g t_1^2}{2} = 0$$

Solve last equation for $t_1$:

$$V \sin(\theta) t_1 - \frac{g t_1^2}{2} = 0$$ $$t_1 \left(V \sin(\theta) - \frac{g t_1}{2}\right) = 0$$ $$t_{11} = 0$$ $$t_{12} = \frac{2V \sin(\theta)}{g}$$

Obviously, we need second solution to use. Put it into equation $x(t_1)$ to determine angle $\theta$:

$$V \cos(\theta) t_1 = 75$$ $$\frac{V \cos(\theta) 2V \sin(\theta)}{g} = 75$$ $$2 \sin(\theta) \cos(\theta) = \sin(2\theta)$$ $$\frac{V^2}{g} \sin(2\theta) = 75$$ $$2\theta = \operatorname{asin}\left(\frac{75g}{V^2}\right)$$ $$\theta = \frac{1}{2} \operatorname{asin}\left(\frac{75g}{V^2}\right)$$

Substitute

$$V = 35; \quad g = 10;$$ $$\theta = \frac{1}{2} \operatorname{asin}\left(\frac{75 \cdot 10}{35^2}\right) \approx 18.88{}^\circ$$

As soon as velocity along $x$ – constant, stone pass middle point at time $\frac{t_1}{2}$. Calculate $y\left(\frac{t_1}{2}\right)$:

$$y \left(\frac {t _ {1}}{2}\right) = V \sin (\theta) \frac {t _ {1}}{2} - \frac {g}{8} t _ {1} ^ {2} = \frac {\left(V \sin (\theta) t _ {1} - \frac {g}{2} t _ {1} ^ {2}\right)}{2} + \frac {g}{8} t _ {1} ^ {2}$$

Recall that

$$V \sin (\theta) t _ {1} - \frac {g}{2} t _ {1} ^ {2} = 0$$

Then

$$y \left(\frac {t _ {1}}{2}\right) = \frac {g}{8} t _ {1} ^ {2} = \frac {g}{8} \left(\frac {2 V \sin (\theta)}{g}\right) ^ {2} = \frac {g}{8} \frac {4 V ^ {2}}{g ^ {2}} \sin^ {2} \left(\frac {1}{2} \mathrm {a s i n} \left(\frac {7 5 g}{V ^ {2}}\right)\right) = \frac {V ^ {2}}{2 g} \sin^ {2} \left(\frac {1}{2} \mathrm {a s i n} \left(\frac {7 5 g}{V ^ {2}}\right)\right)$$

Substitute

$$V = 3 5; \quad g = 5;$$ $$y (t _ {1}) \approx 6. 4 1 > 3$$

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