Question

A ball was thrown with an initial vertical velocity of 20 m/s and an initial horizontal velocity of 30 m/s.

a. What will each component velocity be after one second?

b. What will be the total time of the flight of the ball? Consider that the ball was released from the ground and landed some meters away.

c. Compute the range.

Answer 1

(a)

$$v_x(t=1\ s)=v_{0x}=30\ \dfrac{m}{s},$$ $$v_y(t=1\ s)=v_{0y}-gt=20\ \dfrac{m}{s}-9.8\ \dfrac{m}{s^2}\cdot1\ s=10.2\ \dfrac{m}{s}.$$

(b) Let's first find the initial velocity of the ball from the Pythagorean theorem:

$$v_0=\sqrt{v_{0x}^2+v_{0y}^2}=\sqrt{(30\ \dfrac{m}{s})^2+(20\ \dfrac{m}{s})^2}=36.05\ \dfrac{m}{s}.$$

Then, let's find the launch angle from the geometry:

$$\theta=cos^{-1}(\dfrac{v_{0x}}{v_0}),$$ $$\theta=cos^{-1}(\dfrac{30\ \dfrac{m}{s}}{36.05\ \dfrac{m}{s}})=33.7^{\circ}.$$

Let's first find the time that the ball takes to reach its maximum height:

$$v_y=v_0sin\theta-gt,$$ $$0=v_0sin\theta-gt,$$ $$t=\dfrac{v_0sin\theta}{g}.$$

Finally, we can find the total time of the flight of the ball:

$$t_{flight}=2t=\dfrac{2v_0sin\theta}{g},$$ $$t_{flight}=\dfrac{2\cdot36.05\ \dfrac{m}{s}\cdot sin33.7^{\circ}}{9.8\ \dfrac{m}{s^2}}=4.08\ s.$$

(c) We can find the range as follows:

$$x=v_{0x}t_{flight}=30\ \dfrac{m}{s}\cdot4.08\ s=122.4\ m.$$