Answer to Question #152112 in Physics for Ryan Ebuenga

Question

2. A ball is fired from the origin with an initial velocity of 60.0 m/s at 70.0◦ above the x-axis.
Ignore air resistance.
(a)What are the horizontal and vertical components of the ball’s initial velocity?
(b)Find the maximum height above the ball can reach.
(c) What are the horizontal and vertical components of the ball’s velocity and acceleration at its maximum height?

Answer 1

a) Let's find the horizontal and vertical components of the ball’s initial velocity:

$$v_{0x}=v_0cos\theta=60\ \dfrac{m}{s}\cdot cos70^{\circ}=20.52\ \dfrac{m}{s},$$ $$v_{0y}=v_0sin\theta=60\ \dfrac{m}{s}\cdot sin70^{\circ}=56.38\ \dfrac{m}{s}.$$

b) Let's write the equation of motion of the ball in vertical direction:

$$y=v_0tsin\theta-\dfrac{1}{2}gt^2, (1)$$

here, $v_0 = 60\ \dfrac{m}{s}$ is the initial velocity of the ball, $t$ is the time of flight of the ball, $\theta=70^{\circ}$ is the launch angle, $y$ is the vertical displacement of the ball (or the height) and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the time that the ball takes to reach the maximum height from the kinematic equation:

$$v_y=v_0sin\theta-gt_{rise},$$ $$0=v_0sin\theta-gt_{rise},$$ $$t_{rise}=\dfrac{v_0sin\theta}{g}.$$

Then, we can substitute $t_{rise}$ into the first equation and find the maximum height:

$$y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(60\ \dfrac{m}{s})^2\cdot sin^270^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=162.2\ m.$$

c) Since there is no acceleration in the horizontal direction, the horizontal component of the ball's velocity remains unchanged during the flight. Therefore, the horizontal components of the ball’s velocity and acceleration at its maximum height equals $v_{x}=20.52\ \dfrac{m}{s},$ $a_x=0$, respectively. The vertical component of the ball's velocity at its maximum height equals zero. The vertical component of the ball's acceleration at its maximum height is $a_y=-g$ .

Answer:

a) $v_{0x}=20.52\ \dfrac{m}{s}, v_{0y}=56.38\ \dfrac{m}{s}.$

b) $y_{max}=162.2\ m.$

c) $v_{x}=20.52\ \dfrac{m}{s}, v_y=0, a_x=0, a_y=-g.$