Vo = 136Ft/s[17].
Xo = 136*cos17 = 130 Ft/s.
Yo = 136*sin17 = 39.8 Ft/s.
b. Y = Yo + g*Tr = 0, 39.8 - 32Tr = 0, Tr = 1.244 s. = Rise time.
h = 2.5 + 39.8*1.244 - 16*1.244^2 = 27.25 Ft a5ove gnd.
16*Tf^2 = 27.25, Tf = 1.305 s. = Fall time.
1.244+1.305 = 2.549 s. to hit ground.
c. Dx = Xo*(Tr+Tf) = 331.37 Ft.
Please help me answer this question. I have finished the other two parametric equations on my own, but I am confused as to how to do this one.
4. A ball is hit at an angle of 17¢X. The ball is hit when it is 2.5 feet above the ground and is hit at a velocity of 136 feet per second.
Using the following parametric equations for the projectile motion problem.
x=(v0cos£c)t
y=h+(v0sin£c)t-16t^2
where v0 represents the initial velocity of the object (in feet per second), and £c represents that angle made with the horizontal. The value of h is the initial height of the object.
a. Write the two parametric equations that represent this situation.
b. At what time will the ball hit the ground? Round your answer to the nearest thousandth AND explain/show how you got your answer.
c. How far has the ball traveled when it hits the ground? Round to the nearest tenth of a foot AND show/explain how you got your answer.
d. If a 10 foot fence is positioned 327 feet from where the ball is hit, will the ball make it over the fence? If so, by how many feet does it clear the fence? If not, how far from the top of the fence does the ball hit?
2 answers
d. 16*Tf^2 = 27.25-10 = 17.25.
Tf = 1.04 s.
Dx = Xo*(Tr+Tf) = 130*2.28 = 296.7 Ft.
327-296.7 = 30.3 Ft. from bottom of fence.
d^2 = 30.3^2 + 10^2 = 1019.3, d = 31.9 Ft from top of fence.
Tf = 1.04 s.
Dx = Xo*(Tr+Tf) = 130*2.28 = 296.7 Ft.
327-296.7 = 30.3 Ft. from bottom of fence.
d^2 = 30.3^2 + 10^2 = 1019.3, d = 31.9 Ft from top of fence.
1 answer