Physics- Projectile Motion (10 points)?

Is it really telling you to first calculate the speed?

I *think* it is telling you to write an equation for speed... as a function of time !

I'll show you what that means...

Let V be the initial velocity of the ball.

The horizontal component of that velocity is:

Vh = V×cos(33°)

The horizontal displacement is 21 m, so

(21 m) = V×cos(33°)×t

where t is the total time of flight.

You can rewrite that equation to isolate V.

V = (21 m)/(cos(33°)×t)

Now... in itself, it isn't very useful , but you *can* combine it with another equation.

First, find the vertical component of the initial velocity as a function of V.

Vv = V×sin(33°)

Now write the equation for vertical motion,

(2 m) = V×sin(33°)×t - 4.9t²

and replace V by what you found it to be before.

(2 m) = [(21 m)/(cos(33°)×t)]×sin(33°)×t - (4.9 m/s²)×t²

(2 m) = (21 m)×tan(33°) - (4.9 m/s²)×t²

(4.9 m/s²)×t² = (21 m)×tan(33°) - (2 m)

t² = [(21 m)×tan(33°) - (2 m)] / (4.9 m/s²)

t = √{ [(21 m)×tan(33°) - (2 m)] / (4.9 m/s²) }

t = 1.54 s

Entering this time into the equation for horizontal displacement, you get:

V = (21 m)/(cos(33°)×(1.54))

V = 16.2 m/s

Think of a right triangle, you know you have an angle of 33 degrees. The adjacent side of the 33 degrees will be called "x". The opposite side will be called "y". The last side is the hypotenuse, which we will call "a".

Here is an image for reference: http://www.mathsisfun.com/images/adjacent-opposite...

x = a * cos33

y = a * sin33

We have to add a time factor considering you want to know the time and velocity. Also we need to add in gravity, since it will pull the ball downward.

The velocity we will call "v". This velocity (initial speed) is in the same direction as side "a" (33 degrees from horizontal). "x" is still a distance, so if we substitute "v" for "a", we need to add time; "t".

x = v * cos33 * t

Lastly we need to add gravity. Since gravity only acts in the up/down direction it will only effect the "y".

gravity = g = 9.81 m/s^2. Gravity is an acceleration and since "y" is a distance we need to add a time squared: t^2

y = (v * sin33 * t) - 1/2 * g * t^2

Now you have to solve the two equations. Since "x" and "y" are known, you are left with two unknowns "v" and "t".

So first rewrite x = v * cos33 * t

v = x/(cos33 * t)

Substitute the above equation into the "y" equation.

y = (x/(cos33 * t) * sin33 * t) - 1/2 * g * t^2

y = (x*sin33/cos33) - 1/2 * g * t^2

y = (x*tan33) - 1/2 * g * t^2

1/2*g*t^2 + y = x*tan33

1/2*g*t^2 = x*tan33 - y

g*t^2 = 2*(x*tan33 - y)

t^2 = 2*(x*tan33 - y)/g

t = sqrt(2*(x*tan33 - y)/g

Fill in the values:

t = sqrt(2*(21*tan33 - 2)/9.81) = 1.54032 = 1.54 seconds

you have a time, now you can fill in the "t" to get the velocity "v" (initial speed)

x = v * cos33 * t

v = x/(cos33 * t)

v = 21/(cos33 * 1.54) = 16.2595 m/s = 16.3 m/s

the eq would be s = ut + 1/2at^2 (u being the upward force)

a cricket ball being hit at a speed of 30ms-1 at an angle of 45 degrees how far would it travel?

a = 10ms-2 Sh = tbf

Uv = 45sin30ms-1 Uh = 45cos30ms-1

Sv = 0 t = tbf

t = tbf Sh = Uht

Vv = tbf

0 = 21-213xt + 5t^2 Sh = 30cos45 x 6sin45 = 90

t(21.213 + 5t) (6sin45 is 4.2426)

= 0 & 4.2426

0 = 44t + 5t^2

t = (44 + 5t)

44/5 + 8.8 sec

take gravity as 10 for ease of calculations then multiply the answer by 0.981 at the end to be exact.

as both the cos and sin of 45o are 0.707 this the maximum distance a projectile can go, and come down under power.

if you rework the eq you can find the initial velocity of the football.

x=(ucos33)t

y=(usin33)t-1/2gt2

here x=21

y=2

by solving the above 2 equations you will get

u=16.21m/s

t=1.54s

I'm sorry but I can only solve your question,by finding time first,maybe someone else will be able to.Here's my solution though:

H=1/2gt^2

H=2m,g=-9.8m/s^2,t=?

t=0.64sec

To find initial velocity:

v=d/t

vcos33=21m/0.64sec

v=32.8/cos33=39.1m/s

Hope this helps and again sorry for not being able to find intial velocity first.