A small ball rolls horizontally off the edge of a tabletop that is 1.2 m high. It strikes the floor at a point 1.52m horizontally away from the edge of the table. a) How long is the ball in the air? b) What is the speed at the instant it leaves the table?

I don't understand which equation I would use to solve this problem.

How long does it take to fall 1.2 m?
d= 1/2 g t^2.

solve for time.

Now, that is the same time it took to go 1.52 m horizontally, solve for speed.

wow - so obvious lol, thank you bob ;)

A ball travelling with a speed 10ms^-1 rolls off a horizontal table of 1.0metre high ignoring air resistance,calculate the magnitude of the vertical component of the velocity of the ball as it reach the ground

You're welcome! Don't worry, sometimes problems can seem more complicated than they actually are. It's always important to break them down into smaller steps and equations. If you have any more questions, feel free to ask!

You're welcome! Let's go through the steps to solve the problem.

a) To find out how long the ball is in the air, we can use the equation for vertical motion:

d = 1/2 * g * t^2

where:
d is the vertical distance (1.2 m in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time the ball is in the air (what we are trying to find)

We can rearrange the equation to solve for time:

t = sqrt(2d / g)

Substituting the values:

t = sqrt(2 * 1.2 / 9.8) ≈ 0.49 seconds

So, the ball is in the air for approximately 0.49 seconds.

b) To find the speed of the ball at the instant it leaves the table, we can use the equation for horizontal motion:

v = d / t

where:
v is the velocity (what we are trying to find)
d is the horizontal distance (1.52 m in this case)
t is the time we just calculated (0.49 seconds)

Substituting the values:

v = 1.52 / 0.49 ≈ 3.10 m/s

Therefore, the speed of the ball at the instant it leaves the table is approximately 3.10 m/s.

Remember, it's important to understand the underlying principles and equations in order to solve physics problems. I'll be here to help if you have any more questions!