Question
A cannon is fired horizontally from atop a 40.0 m tower. The cannonball travels 180 m horizontally before it strikes the ground. With what velocity did the ball leave the muzzle?
Okay so this is my work for this problem. please help If I did something wrong. Thanks!
40=1/2(9.8)t^2
180/t
63m/s
Okay so this is my work for this problem. please help If I did something wrong. Thanks!
40=1/2(9.8)t^2
180/t
63m/s
Answers
Your work is correct, here's the explanation.
In a projectile problem, the horizontal (x) motions are independent of the vertical (y) motions.
In the case of a horizontally fired projectile from a given height, the problem is equivalent to a free fall from the given height of 40m PLUS a horizontal uniform velocity travelled during the free-fall time, t seconds.
Free-fall time, t:
Δy = Viy*t+(1/2)a*t²
Here a=-9.8 m/s², V0=0 (horiz.)
giving
t=sqrt(2Δy/g)
=sqrt(2*(-40)/(-9.8))
=2.857s. (approx.)
During 2.857s, the cannonball travelled 180m, so initial velocity
Vix = 180 m /2.857 s = 63 m/s
In a projectile problem, the horizontal (x) motions are independent of the vertical (y) motions.
In the case of a horizontally fired projectile from a given height, the problem is equivalent to a free fall from the given height of 40m PLUS a horizontal uniform velocity travelled during the free-fall time, t seconds.
Free-fall time, t:
Δy = Viy*t+(1/2)a*t²
Here a=-9.8 m/s², V0=0 (horiz.)
giving
t=sqrt(2Δy/g)
=sqrt(2*(-40)/(-9.8))
=2.857s. (approx.)
During 2.857s, the cannonball travelled 180m, so initial velocity
Vix = 180 m /2.857 s = 63 m/s