To win the game, a place kicker must kick a
football from a point 28 m (30.6208 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 18 m/s at
an angle of 48◦
from the horizontal.
The acceleration of gravity is 9.8 m/s
2
.
By how much vertical distance does the ball
clear the crossbar?
Answer in units of m.

Respuesta :

v = 18cos48 m/s  

So time = distance/speed  

= 28/18cos48  

= 2.3247413  

Resolving vertically, now use s = ut + (0.5)a(t^2)  

Where s is the unknown distance (the height above the GROUND)  

u is initial (vertical) speed = 18sin48  

a = -9.8 m/s (negative since we take upwards as positive)  

t = 2.3247... (what we found previously)  

Putting the numbers into the formula gives 4.616126809  

Take away the height of the goalpost (3.05) = 1.566126809  

= 1.57m above the crossbar.

hope this helps :)

We have that the  vertical distance the ball clear the crossbar is

x=1.795

From the question we are told

  • To win the game, a place kicker must kick a football from a point 28 m (30.6208 yd) from  the goal,
  • the ball must clear the crossbar,  which is 3.05 m high
  • the ball  leaves the ground with a speed of 18 m/s at  an angle of 48◦
  • The acceleration of gravity is 9.8 m/s

Generally the  equation for the t  is mathematically given as

[tex]t=\frac{d}{vcos\theta}\\\\t=\frac{28}{18cos48}\\\\t=2.3sec[/tex]

Generally the Newtons equation for the distance  is mathematically given as

[tex]s=uyt+1/2at^2\\\\s=18sin48*(2.3)-\frac{9.8}{2}*(2.3)^2\\\\s=4.845[/tex]

Therefore

The Distance clear will be

[tex]x=s-3.05\\\\x=4.845-3.05[/tex]

x=1.795

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