Question

question 2
a ball is thrown from level ground with a velocity of 20 m/s at an angle of 30° above the horizontal. determine;
a. the time required to reach maximum height in meters
b. the ball’s maximum height in meters.
c. the elapsed time just before the ball returns to the ground in seconds.
d. the ball’s x - coordinate at t = 1.5 seconds in meters.
e. the magnitude of the ball’s velocity at t = 1.5 seconds in m/s.
f. the ball’s direction of travel at t = 1.5 seconds in degrees.
g. is the ball rising or falling at t = 1.5 seconds?

Explanation:

Step1: Find initial vertical velocity

The initial velocity $v_0 = 20$ m/s and the launch - angle $\theta=30^{\circ}$. The initial vertical velocity $v_{0y}=v_0\sin\theta=20\times\sin30^{\circ}=10$ m/s, and the initial horizontal velocity $v_{0x}=v_0\cos\theta = 20\times\cos30^{\circ}=10\sqrt{3}$ m/s.

Step2: Time to reach maximum height (a)

At maximum height, the vertical velocity $v_y = 0$. Using the kinematic equation $v_y=v_{0y}-gt$, where $g = 10$ m/s². Solving for $t$ gives $t=\frac{v_{0y}-v_y}{g}=\frac{10 - 0}{10}=1$ s.

Step3: Maximum height (b)

Using the kinematic equation $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$. At maximum height $t = 1$ s, $y - y_0=10\times1-\frac{1}{2}\times10\times1^{2}=5$ m.

Step4: Time of flight (c)

The time of flight $T$ is twice the time to reach maximum height. So $T = 2t=2$ s.

Step5: x - coordinate at $t = 1.5$ s (d)

Using the equation $x=v_{0x}t$, with $v_{0x}=10\sqrt{3}$ m/s and $t = 1.5$ s. Then $x=10\sqrt{3}\times1.5 = 15\sqrt{3}\approx25.98$ m.

Step6: Vertical velocity at $t = 1.5$ s

Using $v_y=v_{0y}-gt$, with $v_{0y}=10$ m/s, $g = 10$ m/s² and $t = 1.5$ s. So $v_y=10-10\times1.5=- 5$ m/s. The horizontal velocity $v_x=v_{0x}=10\sqrt{3}$ m/s.

Step7: Magnitude of velocity at $t = 1.5$ s (e)

The magnitude of the velocity $v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{(10\sqrt{3})^{2}+(-5)^{2}}=\sqrt{300 + 25}=\sqrt{325}=5\sqrt{13}\approx18.03$ m/s.

Step8: Direction of velocity at $t = 1.5$ s (f)

The direction $\theta=\tan^{-1}(\frac{v_y}{v_x})=\tan^{-1}(\frac{-5}{10\sqrt{3}})\approx - 14.48^{\circ}$. The negative sign indicates that the angle is below the horizontal.

Step9: Rising or falling at $t = 1.5$ s (g)

Since $v_y=-5$ m/s (negative), the ball is falling at $t = 1.5$ s.

Answer:

a. $1$ s
b. $5$ m
c. $2$ s
d. $15\sqrt{3}\approx25.98$ m
e. $5\sqrt{13}\approx18.03$ m/s
f. $\approx - 14.48^{\circ}$
g. Falling