Projectile Motion - A ball launched from a cliff

  • Thread starter samantha.
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  • #1
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Homework Statement


A ball is launched from a 35.0m high cliff with an initial velocity of 47.0m/s at an angle of 39.0 degrees above the horizontal. On the way down it just clears a 2m high wall.
A) How far away is the wall?

Homework Equations


Delta D=Vx*T + 1/2*a*t^2


The Attempt at a Solution



I calculated the Verticle velocity to be 36.53m/s and the horizontal velocity to be 29.58m/s. Using the Velocity in the in the y direction, and the acceleration due to gravity, I used this equation to calculate time;

Df-Di=Vy*t + 1/2a*t^2
4.905m/s^2*T^2+36.53m/s*t-33.0m

Then used the quadratic equation to get 0.813985s

I plugged this into D=Vx*T and got 24.07m

I figured I mixed up my signs somewhere along the long so I re did it and got 8.26s and then got 244.4m.

The correct answer is 256m.. I'm really close but I don't know why I'm 12 off.. I checked everything over.

Thanks!!
 

Answers and Replies

  • #2
PeterO
Homework Helper
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Homework Statement


A ball is launched from a 35.0m high cliff with an initial velocity of 47.0m/s at an angle of 39.0 degrees above the horizontal. On the way down it just clears a 2m high wall.
A) How far away is the wall?

Homework Equations


Delta D=Vx*T + 1/2*a*t^2


The Attempt at a Solution



I calculated the Verticle velocity to be 36.53m/s and the horizontal velocity to be 29.58m/s. Using the Velocity in the in the y direction, and the acceleration due to gravity, I used this equation to calculate time;

Df-Di=Vy*t + 1/2a*t^2
4.905m/s^2*T^2+36.53m/s*t-33.0m

Then used the quadratic equation to get 0.813985s

I plugged this into D=Vx*T and got 24.07m

I figured I mixed up my signs somewhere along the long so I re did it and got 8.26s and then got 244.4m.

The correct answer is 256m.. I'm really close but I don't know why I'm 12 off.. I checked everything over.

Thanks!!
4.905*T^2+36.53*t-33.0m

In this line - I so prefer it without the units in the middle - you have 4.9 positive, and 36.53 positive

One of those is acceleration due to gravity - down, the other is the initial velocity - up, so one of them should be negative. be careful with your sign on the displacement. it is down, but on the other side of the equation????

also given the supplied formula

Delta D=Vx*T + 1/2*a*t^2

why wasn't your first substitution

33 = -36.53*t + 4.905*t2

followed by a re-arrangement and solution.
 
  • #3
15
0
Okay I re did it and fixed the signs

4.906*t^2-36.53*t+33=0

After doing the quadratic equation, I ended up with the same answer that I did my second attempt :/
 

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