A soccer player can kick the ball 29 m on level ground, with its initial velocity at 42 ∘ to the horizontal.?
At the same initial speed and angle to the horizontal, what horizontal distance can the player kick the ball on a 20 ∘ upward slope?
- NCSLv 75 months ago
over level ground,
range x = V²sin(2Θ)/g
29 m = V² * sin84º / 9.8m/s²
V = 16.9 m/s
Now use the trajectory equation:
y = h + x·tanΘ - g·x² / (2v²·cos²Θ)
where y = height at x-value of interest = x*sin20º = 0.3420x
and h = initial height = 0 m
and x = range of interest = ???
and Θ = launch angle = 42º
and v = launch velocity = 16.9 m/s
Substituting values (and dropping units for ease -- x is in meters
0.342x = 0 + x*tan42º - 9.8x² / (2*16.9²*cos²42º)
0 = 0.558384x - 0.031065x²
0 = x*(0.558384 - 0.031065x)
which has a non-trivial solution at
x = 18 m
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- Steve4PhysicsLv 75 months ago
Question has been asked/answered before (but with different values). Take a look to see the method.