physics 11 question?

Question

how do you find the range, for example in a question like this: a rock is projected horizontally at 15 m/s at a height of 34 m above ground what is the range?

Answer 1

Newtons equations are the trick
you are given u = 15 m/s, vertical height = 34 m

bit of a trick in this one. you can ultimately use the equation
s = ut + 1/2 a(t^2), but in this equation s IS NOT 34 metres and you need to know the flight time of the projectile (t).

First note that the projectile is fired horizontally, so is falling toward the ground only under gravity. Determine how long it take to hit the ground.

Use s = ut + 1/2 a(t^2), where in this case u = 0 (its not "pushed" only falling by gravity) so
34 = 1/2 * 9.8 * t^2
t = 2.63 seconds

to calculate the horizontal distance after 2.63 seconds use the same equation:
s = ut + 1/2 a(t^2), but this time u = 15 m/s and we assume horizontal acceleration is 0 (ie constant horizontal velocity with zero friction) and then
s = ut = 15 * 2.63 = 39.45 m

Answer 2

You should be looking for delta x. (or the change in x = distance traveled) Time is the constant between the x and the y values (distance traveled in x and distance traveled in y). SO, solve for time in the y direction, then plug time into the equation to solve for delta x (distance traveled in x direction). Hope this helps.

Answer 3

find the y component of the initial velocity then apply that in one of the kinematic equations using the fact that acceleration = -9.8 m/s^2

Answer 4

d = vt
t = √2*34 m/9.8 m/s^2
d = (15 m/s)√2*34 s^2/9.8
d = 39.512 m

Answer 5

the projectile falls at a constant rate of 32ft per sec. convert ft to meters and solve the problem