Question

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: Initial velocity components

The initial velocity $v_0 = 20$ m/s and the angle $\theta=30^{\circ}$. The initial - vertical velocity $v_{0y}=v_0\sin\theta=20\sin30^{\circ}=10$ m/s and the initial - horizontal velocity $v_{0x}=v_0\cos\theta = 20\cos30^{\circ}=10\sqrt{3}\approx17.32$ m/s. The acceleration due to gravity $g = 9.8$ m/s².

Step2: Time to reach maximum height (a)

At maximum height, the vertical velocity $v_y = 0$. Using the equation $v_y=v_{0y}-gt$, we can solve for $t$. Rearranging gives $t=\frac{v_{0y}}{g}$. Substituting $v_{0y}=10$ m/s and $g = 9.8$ m/s², we get $t=\frac{10}{9.8}\approx1.020$ s.

Step3: Maximum height (b)

Using the equation $v_y^{2}-v_{0y}^{2}=-2gh$. At maximum height $v_y = 0$. Rearranging for $h$ gives $h=\frac{v_{0y}^{2}}{2g}$. Substituting $v_{0y}=10$ m/s and $g = 9.8$ m/s², we have $h=\frac{10^{2}}{2\times9.8}=\frac{100}{19.6}\approx5.106$ m.

Step4: Time of flight (c)

The time of flight $T$ for a projectile launched and landing at the same height is given by $T = \frac{2v_{0y}}{g}$. Substituting $v_{0y}=10$ m/s and $g = 9.8$ m/s², we get $T=\frac{2\times10}{9.8}\approx2.041$ s. But if we consider the symmetry of the motion, the time to reach the ground is the same as the time it took to go up and come back down. Since the time to reach the maximum - height is $t=\frac{v_{0y}}{g}$, the time of flight $T = 2\times\frac{v_{0y}}{g}=2\times1.020 = 2.041\approx2.04$ s.

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

The horizontal motion is a uniform - motion with $x=v_{0x}t$. Substituting $v_{0x}=10\sqrt{3}\approx17.32$ m/s and $t = 1.5$ s, we get $x=17.32\times1.5 = 25.98$ m.

Step6: Vertical velocity at $t = 1.5$ s

Using the equation $v_y=v_{0y}-gt$. Substituting $v_{0y}=10$ m/s, $g = 9.8$ m/s², and $t = 1.5$ s, we have $v_y=10-9.8\times1.5=10 - 14.7=-4.7$ m/s.

Step7: Horizontal velocity at $t = 1.5$ s

The horizontal velocity remains constant, $v_x=v_{0x}=10\sqrt{3}\approx17.32$ m/s.

Step8: Magnitude of velocity at $t = 1.5$ s (e)

Using the Pythagorean theorem $v=\sqrt{v_x^{2}+v_y^{2}}$. Substituting $v_x = 17.32$ m/s and $v_y=-4.7$ m/s, we get $v=\sqrt{(17.32)^{2}+(-4.7)^{2}}=\sqrt{299.9824 + 22.09}=\sqrt{322.0724}\approx17.95$ m/s.

Step9: Direction of velocity at $t = 1.5$ s (f)

Using the formula $\theta=\tan^{-1}(\frac{v_y}{v_x})$. Substituting $v_x = 17.32$ m/s and $v_y=-4.7$ m/s, we get $\theta=\tan^{-1}(\frac{-4.7}{17.32})\approx - 15.15^{\circ}$. In the range of $0 - 360^{\circ}$, $\theta = 360^{\circ}-15.15^{\circ}=344.85^{\circ}\approx344.8^{\circ}$.

Step10: Rising or falling at $t = 1.5$ s (g)

Since $v_y=-4.7$ m/s (negative), the ball is falling.

Answer:

a. 1.020
b. 5.106
c. 2.041
d. 25.98
e. 17.95
f. 344.8
g. falling