How Do You Calculate Projectile Motion from a Cliff?

[fcb]

Homework Statement

A projectile is fired from the top of a 10m high cliff. It has a range of 152m and reaches a maximum height of 8.2m above the cliff top. Determine the time of flight

Is Delta Y =18.2?
Is U_y=10

Homework Equations

Delta Y = U_yt + 0.5 a_yt

The Attempt at a Solution

18.2=10t-4.9t²
4.9t²-10t+18.2

Using the quadratic equation
10+-21.37/9.8
t=3.2

Does that seem right

 

[dreit]

I believe delta y would be 18.2, and your work seems right.

 

[fcb]

Thank you Drelt

 

[rl.bhat]

You have not used the range of the projectile any where.

 

[fcb]

Where would range play a part in the question? Doesn't range (delta)x only matter on horizontal motion?

I mean if you can find time the you can find Ux but I didn't know you needed to use it to find time

 

[rl.bhat]

Since range is given, the projectile is fired with some angle with the horizontal line.

In the problem they have not given uy = 10 m/s.

 

[fcb]

So is the time not 3.2 since I assumed Uy was 10m/s?

Wouldn't I need to know U to be able to find Uy then. And Ux would be range over time. But how would you find time without finding Uy

Could you make Vy=0 and use the equation
Vy^2 = Uy^2 + 2a (delta) Y

 

[rl.bhat]

Could you make Vy=0 and use the equation
Vy^2 = Uy^2 + 2a (delta) Y

Yes. You are right.

Uy = (2*g*y')^1/2.

Ux*t = 152. So t = 152/Ux.

Now -y = Uy*t - 1/2*g*t^2

-y = (2*g*y')^1/2.*152/Ux - 1/2*g*(152/Ux)^2

Solve the quadratic to find Ux, and then find t.

 

[fcb]

I got t= 3.18

Is that right. Maybe I screwed up somewhere along the line.

I'll double check in the morning as I am in Australia and it's getting late

 

[donutz610]

all looks good, just double check the quadratic work in a calculator on n the computer to make sure you didnt make any mistakes.

 

[fcb]

Nah I think there was a mistake. Can Someone confirm with me that t=~1.903

 

[dreit]

Im not sure if I am correct or not, but when I did the problem i got t= 4.642s

 

[dreit]

I used the equation (deltax)= v(initial)t + (.5)(9.8)t^2

Step One) 152=10t+4.9t^2

Step One rearranged for quadratic formula) 4.9t^2+10t-152=0

Got answers 4.62s and -6.683s, but time cannot be negative so i ruled out the second answer

 

[vissh]

hello :)
Let the projectile is projected with velocity 'u' making an angle A with horizontal.
Let us break the motion in 2 parts :-
<1> From point of projection to maximum height of 8.2m above the cliff top :-
{...For a projectile, projected with an angle A and velocity u,time of flight when the projectile fall back to same horizontal is t =(2usinA)/g
Time to reach max height, t_m= (usinA)/g
Maximum height attained H = (usinA)t_m - (1/2)g[t_m]²
........= (1/2)(usinA)²/g
...} Using this for Case 1 :-
8.2 = (1/2)(usinA)²/g
=>(usinA)² =16.4g
Thus,time to complete this part of motion, t₁ =(usinA)/g = (16.4/g)^1/2

<2> From maximum height of 8.2m above the cliff top to end of flight :-
At max height, only horizontal comp of velocity is there.
Thus initial velocity is ucosA in horizontal direction.
Displacement in vertical height = -(8.2+10) = -18.2m
Thus, -18.2 = 0*t₂ - (1/2)g[t₂]²
=> t₂ = (36.4/g)^1/2

Thus, time of complete flight = t₁ + t₂
.......=3.22 s (approx)

Hope this helps :)

 

[fcb]

Briliant Vissh. Answer seems consistent with fellow class mates. You're an absolute legend

 

[fcb]

I used the equation (deltax)= v(initial)t + (.5)(9.8)t^2

Step One) 152=10t+4.9t^2

Step One rearranged for quadratic formula) 4.9t^2+10t-152=0

Got answers 4.62s and -6.683s, but time cannot be negative so i ruled out the second answer

I think they're different component of x and y and thus you can't use them without linking time

 

[vissh]

Hehe no problem buddy ^.^
[But Please don't over praise xD When u will continue practising , you will also be able to think of different ways of approach :) And solve problems quickly :)]