Question

A player initially at rest throws a ball with an initial speed u = 19.5 m//s at an angle theta = sin^(-1) ((12)/(13)) to the horizontal. Immediately after throwing the ball he starts running to catch it. He runs with constant acceleration (a) for first 2 s and thereafter runs with constant velocity. He just manages to catch the ball at exactly the same height at which he threw the ball. Find ‘a’. Take g = 10 m//s^(2). Do you think anybody can run at a speed at which the player ran?

Answer 1

Answer:

a=$15.9\frac{m}{s^{2} }$

Explanation:

We know,

Velocity in vertical direction = Usin∅ = $19.5$ ×$\frac{12}{13} m/s$

Using mathematical identity,

$sin^{2}x + cos^{2}x = 1\\cos^{2}x = 1-sin^{2}x\\cos^{2}x = 1-\frac{12}{13}.\frac{12}{13} \\\\cos^{2}x =1-\frac{144}{169}\\ cosx=\frac{5}{13}$

Velocity in horizontal direction =Ucos∅=$13.5$×$\frac{5}{13}m/s$

According to equation of motion,

In case of constant acceleration,

$s=ut+\frac{1}{2}at^{2}$

$u=0$

$s=\frac{1}{2} at^{2}$

$t=2sec$

$s=2a$

In case of constant velocity,

$v=u+at$

$u=0$

$v=2at$

Range=$2a+2at$

T=time of flight

Range=$2a+2a(T-2)$

Range=$2a+2aT-4a$

Range=$2aT-2a$

Range=$2a(T-2)$

Range=$\frac{u^{2}sin2 }{g}.$∅

$sin2$∅$=2cos$∅$sin$∅

Range=$(19.5)^{2}.2.\frac{12}{13}.\frac{5}{13}$÷$10$

Range=$27m$

T=$\frac{2Usin}{g}.$∅

T=$2$×$19.5$×$12$÷$13$×$10$

T=$3.6 sec$

$2a(T-2)=27$

a=$\frac{27}{2.(3,6-2)}$

a=$5.19\frac{m}{s^{2} }$

• No anyone can't run at the players speed.