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 23.8 m/s at an angle of 47.0° 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

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the ball = V₀ = 23.8 m/s

Angle of the velocity = = 47^o

Initial velocity of the ball in the horizontal direction = V_x0

V_x0 = V₀Cos

V_x0 = (23.8)Cos(47)

V_x0 = 16.231 m/s

Initial velocity of the ball in the vertical direction = V_y0

V_y0 = V₀Sin

V_y0 = (23.8)Sin(47)

V_y0 = 17.406 m/s

Horizontal distance of the crossbar from the initial point = R = 36 m

Time taken by the ball to reach the crossbar = T

There is no force acting on the ball in the horizontal direction, hence the velocity in the horizontal direction is constant.

R = V_x0T

36 = (16.231)T

T = 2.218 sec

Height of the ball at this time = H

H = V_y0T + gT²/2

H = (17.406)(2.218) + (-9.81)(2.218)²/2

H = 14.476 m

Height of the crossbar = h = 3.05 m

Height by which the ball clears the crossbar = H

H = H - h

H = 14.476 - 3.05

H = 11.426 m

The ball the crossbar vertically by a height of 11.426 m