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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 48β—¦
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 Β 

So time = distance/speed Β 

= 28/18cos48 Β 

= 2.3247413 Β 

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

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

u is initial (vertical) speed = 18sin48 Β 

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

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

Putting the numbers into the formula gives 4.616126809 Β 

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

= 1.57m above the crossbar.

hope this helps :)

okpalawalter8

We have that the Β 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 Β the goal,
  • the ball must clear the crossbar, Β which is 3.05 m high
  • the ball Β leaves the ground with a speed of 18 m/s at Β an angle of 48β—¦
  • The acceleration of gravity is 9.8 m/s

Generally the Β equation for the t Β 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 Β 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

For more information on this visit

https://brainly.com/question/21811998

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