A projectile is fired from a cliff 190 feet above the water at an inclination of 45 to the horizontal, with a muzzle velocity of 50 feet per second The height h of the projectile above the water is given by 32x h(x) x 190 (50) where x is the horizontal distance of the projectile from the face of the cliff. Use this information to answer the following. (a) At what horizontal distance from the face of the cliff is the height of the projectile a maximum? (b) Find the maximum height of the projectile. 6705 feel 32 Simplify your answer Type an integer or a fraction. Type an exact answer. (c) At what horizontal distance from the face of the cliff will the projectile strike the water? 167 feet (Round to the nearest foot as needed.)

Step-by-step explanation:

I'm not sure if this problem is physics based or calculus based.  So I used calculus because it just makes more sense to do so.

If the projectile is fired from 190 feet in the air, then h0 in our quadratic will be 190.  Since we are being asked to find both the displacement in the x-dimension and in the y-dimension, we need a bit of physics here as well.  The velocity in the y-dimension is found in

 and the velocity in the x-dimension is found in

Since the sin and the cos of 45 is the same, it's made a bit simpler for us.  The velocity in both the x and the y dimension is 35.35533906 feet per second.

We can use that now to write the quadratic we need to start solving these rather tedious problems.

The position function for the projectile is

I'm going to kind of mix things up a bit, because in order to find the distance in the horizontal dimension that the object is when it's at its max height, we need to first find out how long it takes to get to its max height.  We will first take the derivative of the position function to get the velocity function of the projectile.  The first derivative of the position function is

Remember that the first derivative is the velocity function of the projectile.  You should know from either physics or calculus that at its max height, the velocity of an object is 0 (because it has to stop in the air in order to turn around and come back down).  Setting the velocity function equal to 0 and solving for time will give us the time that the object is at the max height.

I'm going to factor out the -32 to make things easier:

which gives us that at approximately 1.10485 seconds the object is at its max height.  

Moving over to the horizontal distance question now.  The displacement the object experiences in the horizontal dimension is found in d = rt.  We know the horizontal velocity and now we know how long it takes to get to its max height, so the horizontal distance is found in

d = (35.35533906(1.10485) so

d = 39.06 feet  When the object is at its max height the object is a horizontal distance of 39.06 feet from the face of the cliff.  That's a.

Now to find the max height, we will use again how long it took to get to the max height and sub it in for t in the position function.

s(1.10485) = -16(1.10485)^2 + 35.35533906(1.10485) + 190 to get that the max height is

209 feet.  That's b.

Now for c.  We are asked when the object will hit the water.  We know that when the object hits the water it is no longer in the air and has a height of 0 above the water.  Sub in a 0 for s(t) in the original position function and factor to solve for t:

35.35533906t + 190 and solve for t by factoring however you find to be the easiest.  Quadratic formula works great!

We find that the times are -2.51394 seconds and 4.72365 seconds.  Since time will NEVER be negative, we know that the time it takes to hit the water is 4.72365 seconds.

Whew!!!

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer