A field hockey player hits the ball at A with a speed nuA at

A field hockey player hits the ball at A with a speed, nu_A, at an angle of theta with the i-direction as shown in Figure 1. Assuming the problem to be two-dimensional and neglecting dag on the ball, determine nu_A and theta so that the ball reaches at B with a horizontal speed of 12 ms^-1. An aircraft follows the trajectory shown in Figure 2 and accelerates at 0.02 ms^-2. Determine the reaction force vector exerted on the 75-kg pilot at C if the velocity of the aircraft at A is 55 ms^-1.

Solution

horizontal velocity of a projectile doesn\'t change.
(because there is no force in horizontal direction)

so v cos@ = 12 m/s .........(i)

time taken to reach 35 m in horizontal.

time = X / vcos@ = 35 / 12 = 2.92 s

in vertical,

initial vertcical velocity, vy = vsin@

y = 1 m

a = -9.8 m/s^2

using y = ut + at^2 /2

1 = (vsin@)(2.92) - 9.8(2.92^2)/2

1 = 2.29 vsin@ - 41.68

v sin@ = 18.64 .......(i)

(i) / (ii)

tan@ = 18.64 / 12

@ = 57.23 deg .........Ans

v = 12 / cos57.23 = 22.17 m/s ........Ans

2.

distance of AB curve = @ r

where @ is angle subtends by curve in radians.

@ = 30 pi / 180 = pi/6

AB = 10 x pi / 6 =5.236 miles

BC = 16 x (40 pi / 180) = 11.17 km

total distance traveled = (5.236 x 1609 m ) + (11.17 x 1000 m) = 19594.83 m

using v^2 - u^2 = 2ad

v^2 - 55^2 = 2(0.02)(19594.83)

v = 61.72 m/s

along i vector,

a_t = 0.02 (-i) = - 0.02 i

along j ,

centripetal acc., a_c = v^2 / r = 0.238 j

a_c = 0.238 j

Fnet = ma

F = 75 ( - 0.02i + 0.238j)

F = - 1.5 i + 17.86j