A cannonball is projected from the top of a 50m hill at an angle of 50 degrees above the horizontal. If the ball lands in 9s, what is the initial velocity? The answer is 50m/s [50° N of E] but idk how to get there. Thanks.
- WhomeLv 73 weeks agoBest Answer
Ignore air resistance
Use the general equation for position in time with constant acceleration
s = s₀ + v₀t + ½at²
With the origin at the cannon, and up as the positive direction
in the vertical direction the equation looks like
-50 = 0 + vy₀(9) + ½(-9.8)(9²)
vy₀ = 38.54m/s
sinθ = vy₀ / v
v = 38.54 / sin50
v = 50.316
v = 50 m/s
The N of E makes no sense in this context.
I hope this helps.
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- Steve4PhysicsLv 73 weeks ago
The only way to do this is to assume:
a) The cannonball lands on the horizontal ground, not on the side of the hill, so it ends up 50m lower than where it started.
b) The landing point is 50ºN of E of the launch point. The initial velocity’s direction is therefore 50ºN of E (and 50º elevation).
V = initial velocity.
Initial vertical velocity component u = Vsin(50º).
In the vertical direction, the height change (vertical displacement) is s = -50m.
s = ut + ½at² (you might use different symbols)
Take a = g = -9.8m/s²
-50 = Vsin(50º)*9 + ½(-9.8)*9²
. . . = 6.894V – 396.9
V = (396.9 - 50)/6.894 = 50.3m/s (rounded to 50m/s)