Question

A ball is thrown at an angle of 15° to the horizontal with an initial velocity of 20 m/s.

Answer 1

Answer on Question #50008 – Physics – Mechanics | Kinematics | Dynamics

Given

$$\alpha = 20{}^{\circ}$$ $$v_0 = 20 \frac{m}{s}$$

Solution

we obtain:

$$h = \frac{v_0^2 \cdot \sin^2 20{}^{\circ}}{2g} = \frac{400 \cdot \frac{1 - \cos 40{}^{\circ}}{2}}{2 \cdot 9.8} \, \text{m} \approx 100 \cdot \frac{1 - 0.766}{9.8} \, \text{m} = 2.388 \, \text{m}$$

So $h = 2.388 \, \text{m} \approx 2.4 \, \text{m}$

Answer: $h = 2.4 \, \text{m}$

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