Question

A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.2 m/s at an angle of 49.0° to the horizontal.

(a) By how much does the ball clear or fall short (vertically) of clearing the crossbar? (Enter a negative answer if it falls short.)

Answer 1

Given that

Initial Velocity of ball, V0 = 21.2 m/sec at an angle of 49 deg

V0x = V0*cos A = 21.2*cos 49 deg = 13.91 m/sec

V0y = V0*sin A = 21.2*sin 49 deg = 15.00 m/sec

Now Goal is at a distance of 36.0 m, So in projectile motion (there is no acceleration in horizontal direction, so horizontal velocity will remain constant)

Range = V0x*t

t = Range/V0x = 36.0/13.91 = 2.59 sec

So ball will take 2.59 sec to reach the goal horizontally, Now in that ball's vertical position will be

Using 2nd kinematic equation

h = V0y*t + 0.5*ay*t^2

ay = -g = 9.81 m/sec^2

V0y = 16.00 m/sec

t = 2.59 sec,

So

h = 16.00*2.59 - 0.5*9.81*2.59^2

h = 8.54 m

At that time height of ball = 8.54 m, So ball will clear the crossbar.

Ball will clear by = 8.54 - 3.05 = 5.49 m

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