Question

Consider a gun fired from a tower 96ft above the ground. The angle of elevation of the gun is 60 degrees, and its muzzle velocity is 1600 ft/sec (also gravity in this equation is 32 ft/sec^2).
a)When does the bullet hit the ground?
b)Write a parametric equation that models the path of the bullet.
c)What is the range of the gun?
d) find the equation for the bullet based on rectangular coordinates. What type of curve is this?

Answer 1

x = u t
y = yi + vi t - 16 t^2
----------------------------
u = 1600 cos 60 = 800 ft/sec the whole time
so
x = u t

vi = 1600 sin 60 = 1386 ft/s
so
y = 96 + 1386 t -16 t^2
so
when is y = 0 ?
0 = 96 + 1386 t - 16 t^2
16 t^2 - 1386 t - 96 = 0
so
t = 86.7 or -.07 (forget negative root)
so hits ground in 86.7 seconds
do u t for range

for part d use t = x/u, u is constant
y = yi + vi (x/u) - 16 x^2/u^2
or
y = c + b x + a x^2 a parabola