Physics Projectile Question (Last one)?

Question

Jack be nimble, Jack be quick, Jack jumped over the candlestick with a velocity of 5.0 m/s at an angle of 30.0 degrees to the horizontal. Did Jack burn his feet on the 0.25 m high candle?

Here's the problem... I think I need to know the time of the jump to figure out if he would have made it over or not, but I can't figure out how to do it...
Can someone please explain to me how to do this?
Thanks in advance.

Answer 1

And I thought nursery rhymes couldn't get any lamer...curse you, Jack!

All righty...well, you would have to know his distance away from the candle, when he leapt, but let's assume he leapt and landed at the same distance from said candle.

First, let's identify the vertical and horizontal components of his velocity. (This is always so much easier when I can draw a diagram...)

Create a right triangle using the horizontal and vertical components of his velocity, and a hypotenuse of 5 m/s. The angle between the hypotenuse and the ground is 30 degrees.

There are several methods you could use to solve for the horizontal (hereafter referred to as H) and vertical (V) components. I prefer a/sin a = b/sin b = c/sin c. Just because it's easy to remember. So. 5/sin 90 = H/sin 60 = V/sin 30.
sin 90 = 1
sin 30 = 1/2
sin 60 = approx .866

V = 2.5 m/s
H = 4.3 m/s

(Edit: Had to fix this, I accidentally used 4.3 for V, before...)

H will remain the same, through Jack's jump
t = time, in seconds (relative to the start of the jump)
A = -9.8 m/s^2
Vv = -9.8t + 2.5
P (position) relative to the start of the jump:
Pv = -4.9t^2 + 2.5t
Ph = 4.3t

Solve for Vv = 0. (Vertical velocity equals zero) This is when Jack is at the peak of his jump (which we will assume is directly over the candle)

0 = -9.8t + 2.5
-2.5 = -9.8t
t = .25 seconds

Now figure out Pv (vertical position) at .25 seconds. -4.9 (.25^2) + 2.5 (.25)

-.61 + .625

.625 - .61 = .015 meters, at the peak of his jump

Yeah, he's screwed.

Answer 2

both the time of the jump and the starting distance. Simple triangle geometry (you know the "jump" angle between the long arms of the triangle and the "candlestick" length of the "short" arm) will suffice - assuming of course that Jack jumps in straight lines

The time of the jump becomes more important if Jack travels on a ballistic curve (this is suggested by the inclusion of velocity in the question) - he might hit the candle while coming down

Answer 3

It's a poorly worded problem, because Jack CAN guarantee a burn to his feet if he jumps at the wrong distance from the candle.

I assume that you are supposed to assume that when he is at the peak of his jump, the candle is directly below him. Because of this, you can ignore the x-direction, and just find his final position when he is at the peak of his jump Vy = 0. Acceleration is 9.8 m/s^2 of course.

You only need 3 knowns to solve these projectile problems. Since you know initial velocity (don't forget to find only the Y-component), final velocity (Vf = 0), and acceleration (a = -9.8), you don't need time. Just solve for position, and see if it's greater than 0.25 m.

The formula is Vf^2 = Vi^2 + 2ay, and you need to solve for y.

Answer 4

Jack Jumped Over The Candlestick

Answer 5

What you will do is to apply the formula of time of flight and u=5.0m/s, angle=30.0 degree the acceleration due to gravity (g) =9.8m/s...so applying the formula...u will get the value of t

Answer 6

not to hard, can i please have the projectiles weight so i can work out its momentum ect.?

Answer 7

no, he was wearing Nike.