Question

A stone is thrown from ground level at 79 m/s. Its speed when it reaches its highest point is 60 m/s.
Find the angle, above the horizontal, of the stone’s initial velocity.
Answer in units of ◦

Answer 1

A stone is thrown from ground level at $79\mathrm{m / s}$. Its speed when it reaches its highest point is $60\mathrm{m / s}$.

Find the angle, above the horizontal, of the stone's initial velocity.

Answer in units of $\circ$

Solution:

$V_{0} = 79\frac{\mathrm{m}}{\mathrm{s}}$ — initial velocity of the stone;

$V_{1} = 60\frac{\mathrm{m}}{\mathrm{s}}$ — velocity at the highest point;

$\alpha$ — angle, above the horizontal, of the stone's initial velocity;

During flight, acceleration of gravity is directed along the Y-axis, so the horizontal component of velocity does not change. horizontal component from the right triangle ABC:

$$\Delta ABC: \cos \alpha = \frac{V_{x}}{V_{0}}$$ $$V_{x} = V_{0} \cdot \cos \alpha$$

When the stone is located at the top of the trajectory (highest point), the vertical component of the velocity is zero, then all speed - is the horizontal component.

$$V_{1} = V_{x}$$

(2)in(1):

$$V_{1} = V_{0} \cdot \cos \alpha$$ $$\cos \alpha = \frac{V_{1}}{V_{0}}$$ $$\alpha = \arccos \left(\frac{V_{1}}{V_{0}}\right) = \arccos \left(\frac{60\frac{\mathrm{m}}{\mathrm{s}}}{79\frac{\mathrm{m}}{\mathrm{s}}}\right) = 41{}^{\circ}$$

Answer: angle, above the horizontal, of the stone's initial velocity is $41{}^{\circ}$.