Ball Drop Problem: Find Initial Velocity

  • Thread starter Thread starter ArbazAlam
  • Start date Start date
  • Tags Tags
    Ball Drop
Click For Summary
To solve the Ball Drop Problem, the initial velocity of the first ball must be calculated so that both balls hit the ground simultaneously. The equations for the heights of both balls are set up, with the first ball's height equation accounting for an additional 2 seconds of flight time compared to the second ball. There is confusion regarding the correct use of time variables and the signs in the equations, which is crucial for accurate calculations. The first ball's equation must reflect its longer time in the air, while the second ball's equation starts from the moment it is dropped. Correctly establishing these parameters will lead to the solution for the initial speed needed for the first ball.
ArbazAlam
Messages
10
Reaction score
0
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 2.00 s later. Air resistance may be ignored.

(a) If the height of the building is 60 m, what must be the initial speed of the first ball if both are to hit the ground at the same time?

I've done a number of things that have given me very strange(improbable) numbers. Now what I've attempted to do is set up an equation for each of the balls.

y1(t) = -4.9(t+2)2
y2(t) = vot - 4.9t2

I'm not sure where to go from here. I tried setting y1 to 60 and solving to find the time and then plugging it into y2 but that turned out to be incorrect.
 
Physics news on Phys.org
You've got your times reversed and you must watch your plus and minus signs. The second ball is thrown later. If it takes the second ball , which you are calling 1, 't' seconds to reach the ground, the first ball has t + 2 seconds to reach the ground. Also, what is the direction of the displacement?
 

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?
View full post »

Similar threads

Find the Height of a Building by dropping a ball from the top
  • · Replies 7 ·
Replies
7
Views
3K
Two balls, dropped with a delay of ##\Delta t##, meet after rebound
  • · Replies 34 ·
2
Replies
34
Views
2K
Ball Release and Thrown to reach ground at the same time
  • · Replies 1 ·
Replies
1
Views
2K
Find the buildings height for two balls to land @ same time
  • · Replies 1 ·
Replies
1
Views
2K
One ball dropped, another ball launched
  • · Replies 9 ·
Replies
9
Views
2K
Ball is thrown at the edge of the roof of a building
  • · Replies 31 ·
2
Replies
31
Views
4K
Two Balls Dropped: Same Time Impact?
  • · Replies 2 ·
Replies
2
Views
2K
Distance-time graph of a ball throw vertically up from a fixed point
  • · Replies 5 ·
Replies
5
Views
2K
Ball rolling down a slope problem: Find an expression for time taken
  • · Replies 8 ·
Replies
8
Views
5K
Balls dropped and thrown at the same time
  • · Replies 4 ·
Replies
4
Views
4K