A. The time it takes for the projectile to reach its high point is 5 seconds.
B. The horizontal distance the projectile travels in this time is 250 meters.
A projectile is fired at an angle such that the vertical component of its velocity and the horizontal component of its velocity are both equal to 50 m/s
A. Using g = 10 m/s^2, how long does it take for the projectile to reach its high point?
b. What horizontal distance does the projectile travel in this time?
2 answers
Vertical problem:
v = 50 - g t
if g = 9.81
v = 50 - 9.81 t
v = 0 at top
so
t = 50/9.81 = 5.1 seconds to top
Horizontal problem
horizontal distance during that five second rise = 50 * 5.1 = 255 meters
I bet they really want the horizontal distance for the whole time aloft though:
t = 10.2 seconds in the air (fall time = rise time)
u = 50 * 10.2 = 510 meters
v = 50 - g t
if g = 9.81
v = 50 - 9.81 t
v = 0 at top
so
t = 50/9.81 = 5.1 seconds to top
Horizontal problem
horizontal distance during that five second rise = 50 * 5.1 = 255 meters
I bet they really want the horizontal distance for the whole time aloft though:
t = 10.2 seconds in the air (fall time = rise time)
u = 50 * 10.2 = 510 meters
1 answer