Question

An arrow is shot into the air from the top of a castle wall, which has a height of 98 ft, into the fields surrounding the castle. It is shot with the initial velocity of 54 ft/s, and it reaches its maximum height when it is 107 ft from the base of the castle wall.
How many seconds will it take to land?
How far will it land from the base of the castle wall?
Give your answer correct to the nearest tenth of a unit

[HINT: Use the symmetry of the parabola and acceleration
due to gravity g = -32 ft/s^2

Answer 1

Hi there. Do you just want an answer or do you want to learn the material?

Answer 2

i would like to know the material please

Answer 3

to do this use the equations of constant acceleration for vertical component of the motion
and v = dist/time for horizontal component

Answer 4

im in calculus

Answer 5

i need to do it the calculus way

Answer 6

OK, wonderful. Well the first thing to remember is that the arrow is fired at some angle. Let's call it a. And at the instant it leaves the archer's bow, the total velocity can be divided up into two parts: it's vertical velocity, and its horizontal velocity. The way it's divided up between the two depends on the angle. At high angles, almost all the velocity is vertical. At low angles, it's almost all horizontal.

Here's the thing: the horizontal velocity will remain the same for the whole flight, but the vertical velocity will become less and less (under the force of gravity), eventually become negative, then get faster and faster until the arrow strikes the ground.

Does that make sense so far?

Answer 7

uh huh

Answer 8

OK, so first, using trig, what's the equation for the vertical component of the arrow's velocity, in terms of its initial velocity (54) and its firing angle (a)?

Answer 9

is this the physics way?

Answer 10

no it's the calculus way

Answer 11

it's sort of both, physics and calculus go hand in hand, but i'm not assuming any prior knowledge of the linear motion equations as a physics student would

Answer 12

so i would have a(t) = -32
v(t) = -32t+54

Answer 13

OK, so first, using trig, what's the equation for the vertical component of the arrow's velocity, in terms of its initial velocity (54) and its firing angle (a)?

Answer 14

im not sure..

Answer 15

s(t)=-16t^2+54t+C

Answer 16

i have 4.7 sec takes to land is that right?

Answer 17

Hi I finally figured this stupid thing out too, haha

Answer 18

I set up an equation that related x (distance from wall) to y (height above ground), with three unknown constants (a, b, and c):

h(x) = ax^2 + bx + c, I simultaneously solved:

1) h[0] = 98 (the arrow starts out at 98 feet above)
2) h'[107] = 0 (the arrow's vertical velocity when it's 107 feet away is 0)
3) h''[0] = -32 (the arrow's vertical acceleration is -32feet/second squared)

These three equations, when you differentiate and substitute, come out to:

1) c = 98
2) 214a + b = 0
3) 2a = -32

Simultaneously solve and you get a = -16, b = 3424, c = 98.

Answer 19

Now if you want to know where it lands, you solve for 0:

-16x^2 + 3424x + 98 = 0
x = 214.029

So it lands 214.029 feet from the castle wall.

Answer 20

i came up with the formula -16x^2+54X+98 how do i find the distance from that?

Answer 21

OK I just figured out why this problem was giving me so much trouble. It's impossible. The problem is malformed.

If you shoot the arrow straight up, it will reach its maximum height in 54/32 seconds, or about 1.69 seconds. That's the time it takes gravity to stop an arrow shot at 54 m/s. At any angle other than straight up, the arrow will reach maximum height even sooner; that's the maximum time between firing and max height.

However, even if the arrow were moving directly horizontally for 1.69 seconds, it would only go 91.1 feet horizontally -- short of the 107 feet from the castle base this problem claims.

It's impossible to fire an arrow at 54 meters/second at any angle under the influence of 32m/s/s gravity and have it be 107 feet from the castle base at its maximum height.

Answer 22

ok thanks a lot! so i can't figure this problem out even if i try to?

Answer 23

no, the problem is malformed