One projectile is launched from the side of a building which is 20 meters above

Question

One projectile is launched from the side of a building which is 20 meters above the ground. It is launched at 14 m/s at an angle of 28 degrees above the horizontal and lands on the ground after it travels some horizontal displacement (this would be how far it lands horizontally from the base of the building.) Another projectile is launched from the same height at the same speed, but at an angle of 28 degrees below the horizontal and lands on the ground after traveling some horizontal displacement. How much further does the second projectile travel, in meters, compared to the first?

Explanation / Answer

in vertical,

yf - yi = vyi t + ay t^2 / 2

for 1st projectile,

0 - 20 = (14 sin28)t - 9.8 t^2 /2

4.9 t^2 - 6.57 t - 20 = 0

t = 2.80 sec  

horizontal distance., d = (v0x) t

d = (14 cos28)(2.80) = 34.6 m

FOr second projectile.

0 - 20 = (-14 sin28) t - 9.8 t^2 /2

4.9 t^2 + 6.57 t - 20 = 0

t = 1.46 sec

d = (14 cos28)(1.46) = 18.05 m

distance between them = 34.6 - 18.05 = 16.6 m

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