Question

A projectile is fired from a gun near the surface of Earth. The initial velocity of the projectile has a vertical component of 90 meters per second and a horizontal component of 49 m/s. How long will it take the projectile to reach the highest point in its path?
A100.s B20.s C10.s
D5.s

Answer 1

You only need to use the vertical component of velocity to solve this.

Answer 2

Do you know the kinematics equation?

Answer 3

Oh, is that the distance formula?

Answer 4

Sort of, it's more of a quadratic equation that relates position, velocity, and acceleration.

Answer 5

The clue, "near the surface of the Earth" means that the acceleration is -9.81m/s^2

Answer 6

yes but I don't know the position/distance to slove it, right?

Answer 7

There are a couple other clues here. You are trying to solve for time, so you need an equation that has time as a variable, but a 'hidden given' is that the highest point occurs when the velocity is zero.

Answer 8

Another simplifying assumption is that if you neglect air resistance, the projectile will follow a symmetrical parabolic path - meaning it will take the same time to go up as it does to come back down - so another way to solve it is to solve for how long it takes to get back to ground level and divide that time by 2.

Answer 9

The first kinematics formula looks like \(\Delta s = vt+0.5at^2\)
Where 's' is distance, 'v' is velocity, 't' is time, and 'a' is acceleration.
Have you solved quadratic equations before?

Answer 10

yes I've done those before, I'm having trouble sovling it because I'm not sure what the distance is or time :/ but I got 5 somehow, could that be the answer

Answer 11

but thanks for the help man :D

Answer 12

Try this equation then:
\(\Delta v = at\)
The velocity goes from 90 to zero in some time, t, at an acceleration of -9.81 m/s^2.
You can solve for the time that way as well.