Question

A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined ramp which ends 10 feet above a level landing zone. Assume the cyclist maintains a constant speed up the ramp and the ramp is inclined Ao (degrees) above horizontal. With the pictured imposed coordinate system, the parametric equations of the cyclist will be: x(t) = 100t cos(A) y(t) = –16t2 + 100t sin(A) + 10.
Calculate the horizontal velocity of the cyclist at time t; this is the function x'(t) = _______
What is the horizontal velocity if A = 20 degrees?
What is the horizontal velocity if A = 45 degrees? (Four decimal places.)
Calculate the vertical velocity of the cyclist at time t; this is the function y(t) =________.
What is the vertical velocity if A = 20 degrees? (Four decimal places.)
What is the vertical velocity if A = 45 degrees? (Four decimal places.)
The vertical velocity of the cyclist is zero at time ________ seconds.
If the cyclist wants to have a maximum height of 35 feet above the landing zone, then the required launch angle is A = _______ degrees. (Accurate to four decimal places.)

Answer 1

Answer:

Explanation:

The parametric equations of the cyclist are:

$x(t)=100tcos(A)\\\\y(t)=-16t^2+100tsin(A)+10$   (1)

A) The horizontal velocity is the derivative of x(t), in time:

$x'(t)=100cos(A)$

B) For A=20° the horizontal velocity is:

$v_x=x'(t)=100cos(20\°)=93.9692ft/s$

For A=45°:

$v_x=100cos(45\°)=70.7106ft/s$

C) To find the time in which the vertical velocity is zero you first obtain the derivative of, in time:

$v_y=y'(t)=-32t+100sin(A)+10$

Next, you equal the vertical velocity to zero and solve for time t:

$-32t+100sin(A)+10=0\\\\t=\frac{100sin(A)+10}{32}$

D) The maximum height is reached when the derivative of y (height) is zero. You use the previous value of t in the equation (1), equals y to 35. Next, you  solve for t:

$y=35\\\\-16(\frac{100sin(A)+10}{32})^2+100(\frac{10sin(A)+10}{32})sinA+10=35\\\\-\frac{16}{1024}(10000sin^2A+2000sinA+100)+31.25sin^2A+31.25sinA+10=35\\\\-156.25sin^2A-31.25sinA-1.5625+31.25sin^2A+31.25sinA+10=35\\\\-125sin^2A-26.5625=0$