Question

a cricketer strikes the cricket ball. the ball leaves the bat horizontally at 20 m/s. it hits the ground at a distance of 11 m from the point where it was struck. assume that air resistance is negligible. 5a. calculate the time of flight of the ball. 5b. calculate the vertical velocity of the ball as it reaches the ground. 5c. calculate the vertical distance travelled by the ball during its flight.

Explanation:

Step1: Find time - of - flight (5a)

Use horizontal motion formula $x = v_x t$. Since $v_x=20\ m/s$ and $x = 11\ m$, then $t=\frac{x}{v_x}$.
$t=\frac{11}{20}=0.55\ s$

Step2: Find vertical velocity (5b)

Use vertical - motion formula $v_y=v_{0y}+gt$. Initial vertical velocity $v_{0y} = 0\ m/s$, $g = 9.8\ m/s^2$ and $t = 0.55\ s$.
$v_y=0 + 9.8\times0.55=5.39\ m/s$

Step3: Find vertical distance (5c)

Use vertical - motion formula $y=v_{0y}t+\frac{1}{2}gt^{2}$. Since $v_{0y} = 0\ m/s$, then $y=\frac{1}{2}gt^{2}$.
$y=\frac{1}{2}\times9.8\times(0.55)^{2}=1.48225\ m$

Answer:

5a. $0.55\ s$
5b. $5.39\ m/s$
5c. $1.48\ m$ (rounded to two decimal places)