I need help with a few physics problems?

Question

Please provided details, answers, and how you got to your answer.

Problem 1:
A ball is dropped from the top of a building and hits the ground 4.6 seconds later. How high is the building?

Problem 2:
A cannon fires an object with an inital velocity of 40 m/s at an angle of 25 degrees. How long is the object in the air? What is the maximum height of the object? What is the range of the object?

Problem 3:
A projectile remains in the air for 6.0 seconds after it is fired. The horizontal component of its velocity is 100 m/s. How far did the object move forward before it hit the ground? How long after firing did the projectile reach it's maximum height? How high up did the projectile go?



Thank you. I am studying for a test and really need help

Answer 1

1. You probably were given 4 equations to use in problems involving velocity, time, acceleration etc. What you know is time to reach the ground (t = 4.6 s), original velocity (Vo = 0), acceleration (a = g), and you need the height (the variable name could be s, x, y, or d). To find the one that works for your problem, look at the variable involved in each of the 4 equations. For this problem I hope you have one that looks like this
s = Vo*t + (1/2)*a*t^2

Just plug in the numbers.

2. Break the original velocity into horizontal and vertical components.
Vh = 40 m/s*cos25
Vv = 40 m/s*sin25
Need help knowing when it's sin and when cos? Think about this variation in the setup: What if the angle were zero? Vh would be 40 m/s and Vv would be 0. Cos and sin the way I did it will work that way for an angle of 0, so it's the right way for 25 too.

Using Vv, get the time in the air (up with Vv decreasing, Vv reaches zero, down with Vv increasing negatively, splat) using
V = Vv + a*t
this time. If the ground is level, so the distance from the launch pad to the peak is just as far as the distance from the peak to where it lands, it lands with the same speed, but opposite polarity as it started. (This assumes no air resistance.) So in the formula, after a time t, V will equal -Vv and a = -g (minus because Vv is positive up while g acts downward).

How high? Use
s = (Vo + Vf)*t / 2
where Vf is zero (it briefly stops [vertical motion] at the peak before starting to go down), Vo = Vv, and you just found t. This basically calculates the average velocity on the way up and multiplies by time.
How far horizontally? Use
V = s/t
where V = Vh, s is the unknown, and you know t.

3. Distance to landing?
Vh = s/t
s = Vh*t

Time to max height? It spent 1/2 the time climbing and 1/2 the time falling.

How high? You know how much time it spent falling. Use
s = Vo*t + (1/2)*a*t^2
where Vo = 0, a = g (or -g, doesn't matter since Vo = 0).