# AP Physics C Projectile Question

by Musicman
Tags: physics, projectile
 P: 21 A projectile is fired horizontally from a gun that is 63.0 m above flat ground. The muzzle velocity is 250 m/s. (a) How long does the projectile remain in the air? (in seconds) I got 3.96 s (b) At what horizontal distance from the firing point does it strike the ground? ( in meters) I got 19.44 m (c) What is the magnitude of the vertical component of its velocity as it strikes the ground? I got 187 m because v(x) is a constant and v sub zero is zero in the vertical direction i did 63=.5(9.8)t^2
Emeritus
PF Gold
P: 9,782
 Quote by Musicman A projectile is fired horizontally from a gun that is 63.0 m above flat ground. The muzzle velocity is 250 m/s. (a) How long does the projectile remain in the air? (in seconds) I got 3.96 s (b) At what horizontal distance from the firing point does it strike the ground? ( in meters) I got 19.44 m (c) What is the magnitude of the vertical component of its velocity as it strikes the ground? I got 187 m
What value are you using for g?
 Sci Advisor P: 1,253 Part a. is correct, the other two parts are not. Where are you having trouble?
 P: 21 AP Physics C Projectile Question actually 3.58 s is the time
Emeritus
PF Gold
P: 9,782
 Quote by 0rthodontist Part a. is correct, the other two parts are not. Where are you having trouble?
I think he's is off slightly, I get 3.58 seconds.

Edit: Just seen your revised post now Musicman

Perhaps if you show your working for the remaing two questions.
 Sci Advisor P: 882 I assume you want someone to check you answers? For the first part, I used the formula, d = -1/2 at^2 where d is the distance the projectile falls, a is the acceleration due to gravity (-9.81 m/s^2) and t is the time it takes to hit the ground. 63 m = -1/2 (-9.81 m/s^2) * t^2 t = sqrt (63 / 4.905) = sqrt (12.844) t = 3.58 seconds For the 2nd part, I used my answer for time in the first part along with this formula, d = v*t where d is the horizontal distance traveled, v is the constant horizontal velocity (250 m/s), and t is the time (3.58 s). d = 895.96 meters For part (c), I assumed the horizontal component of the velocity was constant, only the vertical component changed from the intial to final states. To calculate the final vertical velocity, I used the following formula, v = a*t where v is the velocity after a constant acceleration (9.81 m/s^2 due to gravity) is experienced for a time t (3.58 s), v = 9.81 m/s^2 * 3.58 s v = 35.12 m/s Then I added the two velocity vectors together (horizontal and vertical components) using the Pythagorean theorem to get the final, overall, velocity's magnitude, v = 252.45 m/s
 Emeritus Sci Advisor PF Gold P: 9,782 Cum on mrjeffy, don't post complete solutions, we can't watch them squirm then!
 P: n/a I get 3.58 seconds too. Anyway: $$t = \sqrt{\frac{128}{g}}$$
 P: 21 but for the 2nd part, wouldnt i use the vertical direction equation x=(Vo)(t)?
Emeritus
 P: n/a $$x_{max} = v_0t$$ $$x_{max} = 250 * 3.58$$ $$x_{max} = 895 m$$