Calculating Free Fall Motion: A Practice Problem with Step-by-Step Solution

AI Thread Summary
The discussion focuses on calculating the time a ball is in free fall and its initial horizontal velocity after being shot off a 35 m cliff. The initial calculations suggest the ball is in the air for approximately 2.02 seconds, leading to an initial velocity of 19.802 m/s. However, there is confusion regarding the use of equations, as the horizontal distance may not directly influence the time of fall. Participants clarify that the vertical motion is independent of horizontal motion, emphasizing the need to focus on vertical displacement to determine time. The conversation highlights the importance of correctly applying physics principles to solve projectile motion problems.
tmc123
Messages
3
Reaction score
0

Homework Statement



The original question is:
A ball is shot horizontally off the edge of a cliff 35 m high. The ball lands 20 m from the base of the cliff.
a. How long was the ball in the air?
b. What was the initial velocity of the ball?

Homework Equations



I believe these would be helpful.
x - Δx = vt - (1/2)at^2
x - Δx = Vot + (1/2)at^2

The Attempt at a Solution



To find time, I have this:
20 = 0t - (1/2)(-9.8)t^2
20 = 4.9t^2
t = 2.02 sec

which I then plugged into the second equation to find Vo

20 = Vo (2.02) + (1/2)(-9.8)(2.02)^2
20 = 2.02 Vo - 20
40 = 2.02 Vo
Vo = 19.802 m/s

This unfortunately appears to be in error, but I am unsure where I went astray. I think it's something about the incorrect equations being used, or my not fully understanding the problem.
 
Physics news on Phys.org
The ball is being shot horizontally with certain velocity, not given.
The ball is "dropped" vertically from rest and traveled distance given.
You can find time traveled by the ball to the base.
 
Hi - just a quick question to clarify. When you say I can find the time by the ball to the base, are you saying that the x is irrelevant (so I should try setting my equation equal to 0 instead of 20?).
 
x is the product of time and velocity. No forces acting horizontally, so acceleration is irrelevant.
y is given with initial velocity and acceleration(since there is force acting on it, gravity). From this you can find the time.
 
Alright. I'll give that a try. Thank you.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top