If a projectile is fired with an Initial speed of v_0 ft/s at an angle alpha abo

Question

If a projectile is fired with an Initial speed of v_0 ft/s at an angle alpha above the horizontal, then Its position after t seconds is given by the parametric equations x = (v_0 cos(alpha))t y = (v_0 sin(alpha))t - 16t^2 (where x and y are measured in feet). Suppose a gun fires a bullet into the a r with an initial speed of 2048 ft/s at an angle of 30 degree to the horizontal. After how many seconds will the bullet hit the ground? s How far from the gun will the bullet hit the ground? () mi What is the maximum height attained by the bullet? () mi

Explanation / Answer

Solution:

Given;

x=(v0 cos )t
y=(v0 sin )t 16t^2

(a) when it hits the ground... we assume y=0, so

y = (v0 sin )t 16t^2 = 0

=> (v0 sin 16t) t = 0

=> t = 0 and t = (v0 sin ) / 16

Given that v0 = 2048 ft/s and = 30degree

t = (2048 sin(30)) /15 = (2048*0.5)/16 = 64 sec

So after 64 sec will the bullet hit the ground.

(b) x = (v0 cos )t gives us the horizontal displacement it travels

so put t = 64 sec

x = (2048 cos(30)) * 64 = (2048*0.866)*64 = 113508 ft = 21.5 mi

(c) To find the max height divide total flight time by 2

t = 64/2 = 32sec,

plugging the time into y=(v0 sin )t 16t^2

y = (2048 sin(30))*(32) - 16(32)^2

y = 32768 - 16384

y = 16384 ft = 3.1 mi

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