Question

We are standing at a distance d=15 m away from a house. The house wall is h=6 m high and the roof has an inclination angle β=30 ∘. We throw a stone with initial speed v0=20 m/s at an angle α= 38 ∘. The gravitational acceleration is g=10 m/s2.
(a) At what horizontal distance from the house wall is the stone going to hit the roof - s in the figure-? (in meters)
(b) What time does it take the stone to reach the roof? (in seconds)

Answer 1

We are standing at a distance $d = 15 \, \text{m}$ away from a house. The house wall is $h = 6 \, \text{m}$ high and the roof has an inclination angle $\beta = 30{}^\circ$. We throw a stone with initial speed $v_0 = 20 \, \text{m/s}$ at an angle $\alpha = 38{}^\circ$. The gravitational acceleration is $g = 10 \, \text{m/s}^2$.

(a) At what horizontal distance from the house wall is the stone going to hit the roof - s in the figure-? (in meters)

(b) What time does it take the stone to reach the roof? (in seconds)

Solution:

The equation of motion for the stone for the X-axis:

$$x: s + d = V_0 t \cos \alpha$$ $$t = \frac{s + d}{V_0 \cos \alpha}$$

The equation of motion for the stone for the Y-axis:

$$y: h + h_1 = V_0 t \sin \alpha - \frac{g t^2}{2}$$

From the right triangle ABC:

$$\tan \beta = \frac{h_1}{s} \Rightarrow h_1 = s \tan \beta$$

(3) in(2):

$$h + s \tan \beta = V_0 t \sin \alpha - \frac{g t^2}{2}$$

(1) in(4):

$$h + s \tan \beta = \left(\frac{V_0 \sin \alpha}{V_0 \cos \alpha}\right) (s + d) - \frac{g}{2} \cdot \left(\frac{s + d}{V_0 \cos \alpha}\right)^2$$ $$h + s \tan \beta = (s + d) \tan \alpha - \frac{g}{2} \cdot \left(\frac{s + d}{V_0 \cos \alpha}\right)^2$$ $$h + s \tan \beta = s \tan \alpha + d \tan \alpha - \frac{g s^2}{2 (V_0 \cos \alpha)^2} - \frac{s d g}{(V_0 \cos \alpha)^2} - \frac{g d^2}{2 (V_0 \cos \alpha)^2}$$ $$s^2 \cdot \frac{g}{2 (V_0 \cos \alpha)^2} + s \left(\tan \beta - \tan \alpha + \frac{d g}{(V_0 \cos \alpha)^2}\right) + h - d \tan \alpha + \frac{g d^2}{2 (V_0 \cos \alpha)^2} = 0$$ $$0.02013 s^2 + 0.4 s - 3.455 = 0$$ $$s = 6.5 \, \text{m}; -26.38 \, \text{m}$$

We need only the positive root of the equation:

$$s = 6.5 \, \text{m}$$

Time to reach the roof:

$$t = \frac{6.5 \, \text{m} + 15 \, \text{m}}{20 \, \text{m/s} \cdot \cos 38{}^\circ} = 1.36 \, \text{s}$$

Answer: a) the stone is going to hit the roof at distance $s = 6.5 \, \text{m}$ from the house wall;

b) time to reach the roof $t = 1.36 \, \text{s}$

Fig. 1.