I would like assistance with the posted below question: A baseball is hit from a height of 4 ft at a 60° angle above the horizontal. Its initial velocity is 68 ft per sec.(a) Write parametric equations that model the flight of the baseball.(b) Determine the horizontal distance, to the nearest tenth of a foot, traveled by the ball in the air. Assume that the ground is level.(c) What is the maximum height of the baseball, to the nearest tenth of a foot? At that time, how far has the ball traveled horizontally?(d) Would the ball clear a 7-ft-high fence that is 100 ft from the batter?

Name: I would like assistance with the posted below question: A baseball is
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: I would like assistance with the posted below question: A baseball is hit from a height of 4 ft at a 60° angle above the horizontal. Its initial velocity is 68 ft per sec.(a) Write parametric equations that model the flight of the baseball.(b) Determ

Hello. Since your question has multiple sub-parts, we will solve first three sub-parts for you. If you want remaining sub-parts to be solved, then please resubmit the whole question and specify those sub-parts you want us to solve. (a) Assume the velocity to be v at any time t and determine the horizontal and vertical components of the velocity. vhorizontal direction=68cos60°vvertical direction=68sin60° Now, determine the distance travelled in the horizontal and vertical direction. In horizontal direction the baseball is moving with constant speed and in the vertical direction the baseball is travelling under constant acceleration. For constant velocity, d=v×tdhorizontal=68cos60°×tdhorizontal=34tSo, x=34tFor constant acceleration=32.17ft/sec2ho=4ftd=ho+vt−12gt2d=ho+vverticaldirection×t−12gt2d=4+68sin60°t−12×32.17t2d=4+68sin60°t−12×32.17t2d=4+58.889t−16.085t2y=4+58.889t−16.085t2Hence, x=34ty=4+58.889t−16.085t2 are the required parametric equations.(b) First, calculate the time of flight t for the baseball. The baseball hits the ground for y = 0. Evaluate t for y = 0. y=04+58.889t−16.085t2=0t=3.727,−0.0667As time cannot be negative thus, t =3.727 sec. Then, determine the horizontal distance for time of flight t. x=34tx=34×3.727x=126.718Hence, horizontal distance travelled by the ball in air=126.72 ft (c) The vertical velocity is zero at maximum height. Differentiate the vertical distance y w.r.t time t and equate it to 0 to determine the time at which the baseball reaches the maximum height. y=4+58.889t−16.085t2dydt=0ddt4+58.889t−16.085t2=058.889−32.17t=0t=58.88932.17t=1.83 sec Then, determine the height y at t = 1.83 sec. y=4+58.889t−16.085t2y=4+58.889×1.83−16.0851.832y=57.89 ftHence, maximum height =57.89 ftNow, determine the horizontal distance for the time t = 1.83 sec. x=34tx=34×1.83x=62.22Hence, horizontal distance is 62.22 ft at 1.83 sec.