HELP physics homework help?

Question

a projectile is fired with an initial speed 75.2 m/s at an angle of 34.5 above the horizontal on a long flat firing range. a)determine the maxim height reached by the projectile
b) the total time in the air (forget this one)
c) the total horizontal distance covered (that is, the range)
d) the veolcity of the projectile 1.5s after firing


please helpp step by step procedure on doing this =(

Answer 1

given: vo=75.2m/s , theta=34.5
a.
vyo=vosin34.5=75.2*sin34.5=42.6m/s
vy=0(final velocity)
vy=vyo-gt=42.6-9.81t=0
t=4.34 sec.
h=vyo.t-(1/2)gt^2
=42.6*4.34-0.5*9.81*4.34^2
=92.5m
c.
vxo=75.2cos34.5=62m/s
x=t.vx=4.34*62=269m
d.
v=vo-gt=75.2-9.81*1.5=60.48m/s

Answer 2

You have to decompose the initialvelocity vector into its horizontal and vertical components. Fortunately, the problem is a 3-4-5 right triangle situation in disguise, where the hypotenuse is the 75.2 m/sec and the horizontal component is 0.8* 75.2 m/sec or 60 m/sec and the vertical component is 0.6*75.2 or 45 m/sec. Anytime you see 34.5 degree angles, it is a tip-off to this.

The time to max height is given by
0= 45- 9.8t or about 4.6 sec.
The max height is given by
y(max)= 45*4.6 - 4.9*(4.6)^2 or about 100 m
To get the range, you do need to know the time in air, which is twice the time to max height or 9.2 sec. The range is 9.2 sec * 60 m/sec or about 550 meters.

Finally to get the velocity at 1.5 sec, you have to compute the velocity vector from its two components. The vertical component is v(t)=vo-a*t or
v(t) = 45-9.8(1.5) m/sec or about 30 m/sec
The horizontal component is unchanged 60 m/sec. So the velocity will be (by right triangle approach) sqrt(30^2+60^2).
The angle will now be tan-1(30/60) or about 27 deg. above the horizon.

Answer 3

The key to these sorts of problems is to use trig to break down the equations of motion into two ; one in the x direction parallel to the ground and in the y direction up and down... You end up having to calculate b) to find c) so I will walk you through that.

So, take the initial velocity. The y component of motion at the beginning is going to be the muzzle velocity of 75.2 m/s times the sine of the angle (is 34.5 degrees or gradient? Usually degrees, but make sure of your units!).

Now you know how fast it is going up.

A quick shortcut here is to equate the kinetic energy up with the gravitational potential energy....

1/2m*(v_y0)^2 = m*g*h

The m's cancel out, sub in for v_y0 (your initial y velocity) and solve for h.

b) The time up will equal the time down, so, if the trip down time is found by solving for t in 1/2*a*t^2= h, then find "t" and double it to find the total time of flight.

c) Assuming no drag on the projectile (which you nearly always do in these sorts of problems and which I have done so far) now that you have the time of flight simply multiply that by the velocity in the x direction, v_x0=cos(angle)*muzzle velocity. d=v*t, right?

d) This is a vector addition problem.. v=(v_x^2+v_y^2)^(1/2). Draw a little picture with the y vector on the end of the x vector to see why this is. (Remember the pythagorean theorem)


v_x is constant and you figured it out in part c. v_y changes according to the very simple equation v_y(t)=v_y0-a*t, where the. Plug in 1.5 seconds for time and find the velocity after 1.5 seconds (make sure 1.5 seconds is less than the total time in the air!) Once you have v_y(1.5 seconds) then plug into the vector addition problem.

Answer 4

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Answer 5

horizontal component= 75.2cos34.5=
62m/s
vertical component=75.2sin34.5=
42.6m/s. So the vertical component is zero at the highest point: Vy=at=9.8t so t=42.6/9.8=4.35s. Highest point=Vyt-4.9t^2=92.6m (1
time up equal time down so total time in air =2[4.35]=8.7s (2 Total horizontal distance=Vxt=269.4m (3
V(1.5)=sqrt[62^2+(42.6-9.8*1.5)^2]=
68m/s (4