Question

Please solve both with complete solutions if you can!

2/71 The quarterback Q throws the football when the receiver R is in the position shown. The receiver's velocity is constant at 10 yd/sec, and he catches passes quarterback desires the receiver to catch the ball 2.5 sec after the launch instant shown, determine the initial speed vo and angle required. when the ball is 6 ft above the ground. If the Uo 7 30 yd Problem 2/71 2/87 A fireworks shell is launched vertically from point A with speed sufficient to reach a maximum altitude of 500 ft. A steady horizontal wind causes a constant horizontal acceleration of 0.5 ft/sec, but does not af- fect the vertical motion. Determine the deviation δ at the top of the trajectory caused by the wind. Wind 500 Problem 2/87

Answer 1

Since the receiver is running at a speed of 10yard/sec so after 2.5 sec he will be 30+25 yards away from the quarterback. So we can say that the football travels 55 yards in 2.5 sec in positive x direction.

The initial height at which quarterback launch the football is 7 feet and the receiver receive it at the height of 6', so net displacement is in the direction of negative y with a magnitude of 1'.

Fig 1.

Since the receiver is running at a speed of 10 yard/sec so after 2.5 sec he will be 30 + 25 yards away from the quarterback. So we can say that the football travels 55 yards in 2.5 sec in positive x direction. The initial height at which quarterback launch the football is 7 feet and the receiver receive it at the height of 6', so net displacement is in the direction of negative y with a magnitude of 1'. \r\n V_x = V_x yard/sec in +ve x-direction V_y = V_y feet/sec in +ve y-direction acceleration due to gravity = 32 feet/sec^2 in -ve y - direction. Eq. 1 V_x times 2.5 = 55 or, V_x = 55/2.5 = 22 yard/sec Eq. 2 V_y times 2.5 -32/2 (2.5)^2 = -1 {u t + a/2 t^2 = S or, V_y = 39.6 feet/sect or, 1 feet = 0.33 yard so 39.6 feet = 13.2 yard