Physics Question! (Basic Physics)?

Question

A quartback throws the football to a stationary receiver who is 31.5 m down the field. If the football is thrown at an initial angle of 40 degrees to the ground, at what initial speed must the quartback throw the ball for it to reach the receiver? What is the ball's highest point during its flight?


Please show your work so I'll have an idea of what is going on.

Answer 1

Let the minimum velocity be V
Therefore,
V*sin40 is the vertical component. The time the ball would stay in the air with the above vertical component velocity = 2 *V*Sin40/g

V*Cos40 is the horizontal component of the ball. Therefore, in the above timeframe, the ball should have travelled enough horizontally to reach the receiver.

So, V*Cos40* (2*V* Sin40/g) = 31.5
Or, (V^2)* Sin80/g =31.5
Or, (V^2) = 31.5 *9.8/(0.9848)
= 313.4
Or, V = sqrt (313.4) = 17.7 m/s

Answer 2

Let x=horizontal component of speed. Therefore
xtan40=vertical component of speed.

speed =distance/time, where speed is the horizontal component x, distance equals the horizontal distance of 31.5, and t is the time for the ball to reach the qback. Substitute known values:

x=31.5/t Equation (1)

v=u+at where v=final vertical component of the speed at the top of the flight of the ball, u= the initial vertical component of the speed, a is the acceleration of gravity, and t equals the time to reach the top of the flight. Substitute known values:

0=xtan40+(-9.8)t Equation (2)

Now pls note that t in equation (2) has to be doubled to make it equal to the t in equation (1). That's because this t is just the time it takes the ball to reach the top. It will take an equal amount of time t for the ball to reach the ground. So we have to change our equation to:

0=xtan40+(-9.8)t*2 Equation (2)

Substitute in equation (2) the value of x in equation (1):

0=31.5tan40/t-19.6t . Multiply both sides by t:
0=31.5tan40-19.6t^2 Transpose 19.6t^2 to the left.
19.6t^2=31.5tan40 Divide both sides by 19.6
t^2=31.5tan40/19.6
=31.5*0.839/19.6
=1.348
t=(1.348)^1/2
=1.16sec

Substitute this value in (1)

x=31.5/1.16
=27.2m/s
Initial speed =x/cos40
=27,2/0.766
=35.4m/s

Highest point is the distance from ground to the top. Use the formula:

v^2-u^2=2as. Substitute known values:

0-(xtan40)^2=2(-9.8)s

(27.2*0.839)^2=19.6s
519.8=19.6s
s=519.8/19.6
=26.5meters

Note that 9.8 is negative because the ball is going opposite the direction of gravity.

Pls check my calculations. This problem is not only basic physics but also basic trigonometry.

Answer 3

You cannot do modern physics without a solid grounding in classical physics. Modern physics is post Michealson Morley experiment, Lorentz-Fitzgerald and Einstein. Modern Physics tries to resolve some of the paradoxes that started cropping up in classical physics at the extremes. It succeeds more or less. Classical physics is usually a lot more intuitive and therefore easier. If you are going into biology and wish to get serious about it you will need modern physics for some of the biochemistry. Both are fun.

Answer 4

use these formulas

v=u+at
x=0.5(u+v)(t)
x=ut+0.5(a)(t)(t)
v(v)=u(u)+2(a)(x)

use it in x and y direction.
TO GET component velocity

v(x)=v cos (theta)
v(y)=v sin (theta)

do ur homework okey?,,
and gravity acceleration is 9.81 m/(s)(s)

Answer 5

when did they start measuring football fields in metric?????