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Pre-AP physics help

  • #1
i need help with this problem:

A projectile is shot from the edge of a cliff h = 105 m above ground level with an initial speed of v0 = 125 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-39.
3_39alt.gif

Figure 3-39.

(a) Determine the time taken by the projectile to hit point P at ground level.
____ s
(b) Determine the range X of the projectile as measured from the base of the cliff.
____ km
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
____ m/s (horizontal component)
____ m/s (vertical component)
(d) What is the the magnitude of the velocity?
____ m/s
(e) What is the angle made by the velocity vector with the horizontal?
____ ° (below the horizontal)


I think if somebody could get it started, maybe just get one of the answers, i could figure out the rest, i just cant figure out how to get it started.
 
Last edited:

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
17
List out the data you've been given.

What equations can you use for these kinds of problems ?

Which of these equations relates time (part a) to other quantities in the given data ?

What do you know about resolution of vectors along different directions ?
 
  • #3
193
0
k first you must set the equation into horizontal and vertical components then work from there. Then find the vertical and horizontal velocity.
 
  • #4
17
0
cdhotfire has the golden tip. Split the vector into Vx and Vy and then look at how these change during the fall.
 
  • #5
193
0
yup yup, but i should have said first to find x and y initial velocity, then seperate. :smile:
 
  • #6
so, in other words,
[tex]V_{y} = 125sin(37) \approx 75.23[/tex]
[tex]V_{x} = 125tan(37) \approx 94.19[/tex]

but i dont know what to do with that, and thats only till it gets back down level with the cliff, how would i get the rest?

another question, if something is at constant speed is acceleration 0 or 1?
 
Last edited:
  • #7
310
3
Use the equations of motion.
 
  • #8
here's what i've gotten:
[tex]V_{x} \approx 99.83[/tex]
[tex]v_{y} \approx 75.23[/tex]
used [tex]V_{y}[/tex] to get time:
[tex]V_{f} = V_{i} + at[/tex]
[tex]0 = 75.23 + (-9.8)t[/tex]
[tex]t = 7.68[/tex]
double it to get the real time
[tex]t=15.35[/tex]
now, to get the distance:
[tex]vt = d[/tex]
[tex]99.83(15.35) \approx 1532.39[/tex]
ok so far i got everything i need for until the cannonball comes back level with the cliff
now, dropping below the top of the cliff:
vertical component:
[tex]V_{f}^{2} = V_{i}^{2} + 2ad[/tex]
[tex]V_{f}^{2} = -75.23^{2} + 2(9.8)(105) \approx 7717.55[/tex]
[tex]V_{f} \approx -87.85[/tex]
thats the answer for second part of (c)
first part of (c) would be 99.83 since the horizontal velocity is constant.
now to get the time:
[tex]V_{f} = V_{i} + at[/tex]
[tex]87.85 = 75.23 + 9.8t[/tex]
[tex]t = 1.29[/tex]
so the total time would be
[tex]t = 1.29 + 15.35 = 16.64[/tex]
now, to get the distance
[tex]vt = d[/tex]
[tex]d = 99.83(1.29) \approx 128.78[/tex]
total distance:
[tex]1532.39 + 128.78 = 1661.71[/tex]
but its gotta be in km so
[tex]1661.71/1000 = 1.66[/tex]
now, to get (d) using pythagorean theorum:
[tex]-87.85^{2}+99.83^{2} \approx 17683.65[/tex]
[tex]\sqrt{17683.65} \approx 132.98[/tex]
now, to get (e):
[tex]arcsin(\frac{-87.85}{132.98}) \approx -41.35[/tex]

completed.
 
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  • #9
211
0
Here's an easier way to find the time:

First, find the time for the ball to reach the top of its projectile (which you have). Next, instead of doubling it, find the total time for it to touch the ground (distance = vertical distance it travelled to get to the top of its projectile + height of the cliff) using d=V(initial)squated + 1/2 at(squared)
initial velocity is zero

To find the horiz distance, remember that the horiz speed is constant
 
  • #10
211
0
Sorry, gave you the wrong formula.
d=Vt + 1/2at(squared)
 
  • #11
tried that and still came out with same distance, i think my [tex]V_{x}[/tex] is wrong, but i dont know what else it could be.
 
  • #12
found my mistake, was supposed to use cos instead of tan for [tex]V_{x}[/tex] now ive got everthing right, thanks alot to everybody for all the help
ill fix my post above too.
 
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  • #13
193
0
np man, for those problems just place them in diff components all youll be alright, for the rest we all make mistakes so gl with that. :smile:
 
  • #14
^ bump

I got a question about part c

I was trying to get the vertical components of this velocity
by using this formula


[tex](V_2)^2 = (V_1)^2 + 2a \Delta d

= (105 Sin 37) + 2(-9.8m/s^2)(-125m) [/tex]

[tex] V_2 =\sqrt{2513.190577 m^2}
=50.13173224 m [/tex]

but than the right answer is
-79.9 m/s
using a different kinematic formula (a = change in velocity/ change in time)
 
  • #15
uh, can anybody help me there above?
sorry for bumping it again
 

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