question 2\na ball is thrown from level ground with a velocity of 20 m/s at an angle of 30° above the…

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