A baseball (a projectile) is launched from a height of 2 meters above the ground

Question

A baseball (a projectile) is launched from a height of 2 meters above the ground at a speed of 34 m/s at an angle of 22 degrees above the horizontal. The ball is directly hit a fielder who will catch the ball and the fielder is initially 91 meters from point where the baseball is hit (home plate). If his reaction time is 0.40 seconds (time before he starts to move), what is the minimum speed he much run, in m/s, such that he catches the ball 2 meters above the ground (same height as it was hit)?

Explanation / Answer

here,

intial speed of ball , u1 = 34 m/s

theta = 22 degree

as the fielder catches the ball at the same height,

the horizontal range of ball , x1 = u^2 * sin(2*theta)/g

x1 = 34^2 * sin(2*22)/9.81 m

x1 = 81.9 m

the time of flight , t1 = 2 * u * sin(theta)/g

t1 = 2 * 34 * sin(22)/9.81 s

t1 = 2.5966 s

for the fielder

let the speed be v

horizontal distance travelled = v * ( t1 - reaction time)  

( 91 - 81.9) = v * ( 2.5966 - 0.4)

v = 4.14 m/s

the speed of fielder is 4.14 m/s

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