# 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?

### 2 Answers

- NCSLv 75 months ago
over level ground,

range x = V²sin(2Θ)/g

29 m = V² * sin84º / 9.8m/s²

solves to

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.