Two soccer balls are kicked from an elevated cliff down onto a beach, one is kic

Question

Two soccer balls are kicked from an elevated cliff down onto a beach, one is kicked at a 25 degree and the other is kicked at a 60 degree with both launch speeds being 40 m/s. The cliff is 25 m. high, which one, if either will have a greater speed at impact on the ground? Solve using only equations from Two-dimensional motion, constant acceleration. Two-dimensional motion, constant acceleration Delta x = 1/2(v_ix + v_fx)t Delta x = v_ix t + 1/2 a_x t^2 v_fx = v_ix + a_xt v^2_fx = v^2_ix + 2a_x Delta x Delta y = 1/2(v_iy + v_fy)t Delta y = v_iy t + 1/2 a_y t^2 v_fy = v_iy + a_yt v^2_fy = v^2_iy + 2a_y Delta y

Explanation / Answer

Horizontal component of velocity is constant for both the angles

For theta = 25 deg

So Vx = Vox = 40*cos(25) = 36.25 m/sec

Vy = Voy - g*t

Voy = 40*sin(25) = 16.9 m/sec

Using

-y = Voy*t - 0.5*g*t^2

-25 = (16.9*t)-(0.5*9.8*t^2)
t = 4.56 sec

so vy = Voy-gt = 16.9-(9.8*4.56) = -27.788 m/sec

so for angle theta = 25 deg

speed at the impact on the ground is v = sqrt(36.25^2+27.788^2) = 45.67 m/sec

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for theta = 60 deg

vx = Vox = 40*cos(60) = 20 m/sec

-y = (voy*t0-(0.5*g*t^2)

-25 = (40*sin(60)*t)-(0.5*9.8*t^2)

t = 7.72 sec

then vy= voy - (g*t)

vy = (40*sin(60)) - (9.8*7.72) = -41 m/sec

so speed on the ground is v = sqrt(20^2+41^2) = 45.62 m/sec

so both the bal have same speed at the impact on the ground

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