Question

To win the game, a place kicker met kick a

football from a point 35 m (38.276 yd) from
the goal, and the ball must clear the crombar,
which is 3.05 m high. When kicked, the ball
leaves the ground with a speed of 20 m/s at
an angle of 45.9 from the horizontal.
The acceleration of gravity is 9.8 m/s²
By how much vertical distance does the ball
clear the crowbar?
Answer in units of m.

Answer 1

2.08 m (2 d.p.)

Step-by-step explanation:

Constant Acceleration Equations (SUVAT)

`\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+(1)/(2)at^2\\\\ s&=\left((u+v)/(2)\right)t\\\\v^2&=u^2+2as\\\\s&=vt-(1)/(2)at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^(-1)$\\\\$v$ = final velocity in ms$^(-1)$\\\\$a$ = acceleration in ms$^(-2)$\\\\$t$ = time in s (seconds)\end{minipage}}`

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

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When a body is projected through the air with initial speed (u), at an angle of θ to the horizontal, it will move along a curved path.

Therefore, trigonometry can be used to resolve the body's initial velocity into its vertical and horizontal components:

• Horizontal component of u (x) = u cos θ
• Vertical component of u (y) = u sin θ

Since the projectile is modeled as moving only under the influence of gravity, the only acceleration the projectile will experience will be acceleration due to gravity.

If the ball is kicked at an initial velocity of 20 m/s from flat ground at an angle of 45.9° then:

• Horizontal component of u = 20 cos 45.9°
• Vertical component of u = 20 sin 45.9°

Resolving horizontally

The horizontal component of velocity is constant, as there is no acceleration horizontally.

Resolving horizontally, taking → as positive:

`s=35 \quad u=20 \cos 45.9^(\circ) \quad v = 20 \cos 45.9^(\circ) \quad a=0`

`\begin{aligned}\textsf{Using} \quad s&=\left((u+v)/(2)\right)t\\\\35&=\left((20 \cos 45.9^(\circ)+20 \cos 45.9^(\circ))/(2)\right)t\\35&=(20 \cos 45.9^(\circ))t\\t&=(35)/(20 \cos 45.9^(\circ))\\\implies t&=2.51468288...\; \sf s\end{aligned}`

Therefore, the time is takes for the ball to reach the crossbar is 2.51 s.

Resolving vertically

Acceleration due to gravity = 9.8 ms⁻²

Resolving vertically, taking ↑ as positive:

`u=20 \sin 45.9^(\circ) \quad a=-9.8 \quad t=2.51468288...`

`\begin{aligned}\textsf{Using} \quad s&=ut+(1)/(2)at^2\\\\s&=(20 \sin 45.9^(\circ))(2.51468288...)+(1)/(2)(-9.8)(2.51468288...)^2\\s&=36.1171982...-30.9857870...\\\implies s&=5.131411132...\; \sf m\end{aligned}`

Solution

To find by how much the ball clears the crossbar, subtract the height of the crossbar (3.05 m) from found the vertical height of the ball at 2.51 s:

`\implies 5.13141132...-3.05=2.08\; \sf m\;(2 \: d.p.)`

Therefore, the ball clears the crossbar by 2.08 m (2 d.p.).