Dr. Storr challenged a student to kick a soccerball across a 42, 900.0 mm wide c

Question

Dr. Storr challenged a student to kick a soccerball across a 42, 900.0 mm wide canyon is at the same height on both sides. The student kicked the ball with all his He strength which was placed at the edge of the canyon and achieved an initial velocity m/s at an angle of 39.0 degree. Perform the calculations to see if the ball makes it to the other side. How much or long of the canyon does it travel? Calculate the time of flight of the ball in seconds. Determine the maximum height of the ball in meters.

Explanation / Answer

PROJECTILE

along horizontal

initial velocity vox = vo*costheta

ax = 0

from equation of motion

x-Xo = vox*T+ 0.5*ax*T^2

x-X0 = vo*costheta*T

T = (x-X0)/(vo*costheta)......(1)

along vertical

voy = vo*sintheta

acceleration ay = -Eq/m = 6.4*10^5*1.6*10^-19/(1.67*10^-27) = -6.13*10^13 m/s^2

from equation of motion

y-y0 = voy*T + 0.5*ay*T^2

y-y0 = (vo*sintheta_0*(x-x0))/(vo*costheta_0) + (0.5*ay*(x-x0)^2)/(vo^2*(costheta_0)^2)

y-y0 = (x-X0)*tantheta_0 + ((0.5*ay*(x-X0)^2)/(vo^2*(costheta_0)^2))

sinc the height is same on both sides

y-y0 = 0

0 = 42.9*tan39 - (0.5*9.8*42.9^2/(v0^2*(cos39)^2))

v0 = 20.73 m/s <<<------------answer

-------------

(b)

T = 2*v0*sintheta/g

T = 2*20.73*sin39/9.8

T = 2.66 s

--------

ymax = v0^2*(sintheta)^2/2g

ymax = (20.73^2*(sin39)^2)/(2*9.8) = 8.7 m <<<<-----answer

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