Question

Please answer with a full solution. Thanks!

A Baseball is hit when it is 2.7 feet above the ground. It leaves the bat with n initial speed of 150 ft/sec making an angle of 18 ∘ with the horizontal. Assume a drag coefficient k=6.12. find the range and the flight time of the ball For projection with linear drag x = x 0 + v 0 k ( 1 − e − k t ) c o s α y = y 0 + v 0 k ( 1 − e − k t ) s i n α + 8 k ( 1 − k t − e − k t ) Where k is the drag coefficient and v_0 and \alpha are the projectiles initial speed and the lauch angle, and g is the acceleration of the gravity (32ft/sec^2) a. Flight time=2.93 sec; range=351.6 feet b. Flight time=2.81 sec; range=340.3 feet c. Flight time=3.16 sec; range=362.5 feet d. Flight time=2.01 sec; range=331.6 feet

Answer 1

projectile motion

x = x_o + v_ok ( 1-e^-kt) Cos

y = y_o + v_ok ( 1-e^-kt) Sin + 8k (1- kt -e^-kt )

y_o = 2.7 ft, initial height

k = 6.12 - drag co-efficient.

v_o = 150 ft/s - launch velocity

= 18⁰ launch angle

The projectile will initially raises height , reaches max. height when its vertical velocity =0 and then descends it hits the ground at y=0, 2.7 ft below its launch pos.

initial x-pos =0, we take the horizontal launch pos. as 0 reference.

the best way to determine the flight time and the range is to plot the y(t) and x(t) curves , when y(t) crosses the x-axis (t) , is the time of flight. x(t) value for the flight time is the range of the projectile,as it hits the ground.

The plots are made in excel.

flight time t = 1.12 s and range R = 872 m