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. (These are the parametric equations for the motion of the stunt cyclist.)If the cyclist wants to have a maximum height of 35 feet above the landing zone, then the required launch angle is A=

Answer 1

Answer:$A=23.57^{\circ}$

Explanation:

Given

velocity=100 ft/s

height of landing zone=10 ft

Equation of $x(t)=100 t\cos (A)$

$y(t)=-16t^2+100t\sin (A)+10$

Maximum height=35 feet

at maximum height

$\frac{\mathrm{d} Y}{\mathrm{d} t}=0$

$\frac{\mathrm{d} x}{\mathrm{d} t}=-32t+100\sin A$

$t=3.125\sin A$

At $t=3.125\sin A$

$Y(t)=35=-16\times (3.125\sin A)^2+100\times 3.125\sin A\times \sin A+10$

$312.5\sin^2 A-156.25\sin^2 A=25$

$sin^2 A=0.16$

$A=23.57^{\circ}$

Answer 2

To have a maximum height of 35 feet above the landing zone, the angle of launch required is approximately 23.58°

The reason why the above angle is correct is given as follows:

The given parameters are;

The speed with which the cyclist is travelling, v = 100 ft./sec (constant speed)

Height of the end of the ramp above ground, h = 10 feet

Angle of inclination of the ramp = A₀

Parametric equation for the cyclist are;

x(t) = 100·t·cos(A)

y(t) = -16·t² + 100·t·sin(A) + 10

Required;

To find the required launch angle for a maximum height of 35 feet

Solution:

At maximum height, $\dfrac{d(y(t))}{dt} =y'(t) = 0$

y'(t) = -32·t + 100·sin(A)

At maximum height, y'(t) = 0 = -32·t + 100·sin(A)

Therefore;

$At \ maximum \ height \ t = \dfrac{100 \cdot sin(A)}{32}$

Making sin(A) the subject also gives;

$At \ maximum \ height \ sin(A)= \dfrac{32\cdot t}{100} = \dfrac{8\cdot t}{25}$

From which we have;

$Maximum \ height \ y(t) = -16 \cdot t^2 + 100\cdot t \cdot \dfrac{8 \cdot t}{25} + 10= 16 \cdot t^2 + 10$

When the height is 35 feet, we get;

$Maximum \ height \ y(t) = 35 = 16 \cdot t^2 + 10$

Therefore;

$For \ a \ maximum \ height \ of \ 35 \ feet, \ t =\sqrt{\dfrac{35 - 10}{16} } = \dfrac{5}{4}$

Which gives;

$At \ maximum \ height \ t = \dfrac{5}{4} = \dfrac{100 \cdot sin(A)}{32}$

$sin(A)= \dfrac{5 \times 32}{4 \times 100} = \dfrac{2}{5}$

$\angle A =arcsin \left( \dfrac{2}{5} \right) \approx 23.58 ^{\circ}$

Therefore, if the cyclist wants to have a maximum height of 35 feet above the landing zone, then the required launch angle, A ≈ 23.58°

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