Kicker...?

Question

A place kicker must kick a football from a point 32.7 m (= 35.8 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.6 m/s at an angle of 53.7° to the horizontal. By how much does the ball clear or fall short of clearing the crossbar (enter positive values if the clears and negative values if it fall short)? Also, What is the vertical velocity of the ball at the time it reaches the crossbar? Enter positive values when it is still rising and negative values when it is falling.

Answer 1

Use the initial speed and angle to find the initial horizontal and vertical speeds. The speed times the cosine of the angle is the horizontal speed, V_x, while speed times sine is the vertical speed, V_y. Then, assume a constant horizontal speed to determine how long it takes to travel 32.7 m to the bar by taking 32.7 / V_x. Use the time, t, along with downwards acceleration due to gravity, to calculate the vertical position at that time as y = V_y*t - 0.5*g*t^2. You will be able to easily see if the ball is above or below the 3.05 m bar. You can also calculate the vertical velocity as V_y - gt to see if the velocity is upwards or downwards.

Answer 2

usually these types of ballistics are most easily done by separating the horizontal (x) component and vertical (y) component of the projectile

the horizontal initial velocity is:
cos (53.7 deg) * 21.6 m/s = 12.8 m/s

that means the ball is traveling at 12.8 m/s toward the goal

the goal is 32.7 m away

so, 32.7 m/ 12.8 m/s = 2.55 s

meaning it takes 2.55 seconds for the ball to travel that horizontal distance to the goal

now, how high is the ball at this point?

well, the vertial intial velocity of the ball is:
sin(53.7 deg) * 21.6 m/s = 17.4 m/s

so the ball starts out going up at a speed of 17.4 m/s

it is being accelerated downward by gravity for the whole flight
that acceleration will slow its upward speed until it starts it falling downward

the height of the ball at anytime of its flight is given by:

h=1/2 at^2 +vt (or the height is equal to the initial one half the acceleration times time squared plus the initial velocity times time) in this case that is:

h at time t=2.55 sec (when the ball is moving past the goal)
h= 1/2 (-9.8 m/s^2) * (2.55 s)^2 + 17.4 m/s * 2.55 s
h=.5*-9.8*2.55^2 m + 17.4*2.55 m
h= -31.9 m + 44.37 m =12.1 m

so, the ball is 12.1 m high when it moves horizontally past the goal

the goal is 3.05 meters tall

you can do the final arithmetic I bet

ALSO, you better check my math. It is easy to make an arithmetic error typing into these boxes and using the sorry windows calculator. Hope this gives you a feel for how these work.