Question

A projectile is fired from ground level over ground level with initial vertical velocity 20 m/s and horizontal velocity 30 m/s. using g = 10m/s^2, find the distance from launching to ending point.

Answer 1

Hi Sex dx!

Answer 2

How long will it take for the projectile to hit the ground again?

Answer 3

You don't know? Do you want an explanation?

Answer 4

Use the vertical velocity to find the time of flight, using
\[\Large v_f=v_i+at\]
where vf is final vertical velocity (it will be equal in magnitude to the initial vertical velocity, but in the opposite direction, so -20m/s, keep in mind a will be -10m/s/s, vi will be +20m/s)

Use that time of flight to find distance travelled:
\[\Large d=v_x t\]
where vx is the horizontal velocity.

Answer 5

I got a launch angle of 33.69 degrees and then I used Vy - VoSin33.69 - gt to find t=.34s then I plugged that into X-Xo-30sin33.7(.34) to get 5.65m but the correct answer is 120m

HELP

Answer 6

Follow the steps I gave above. You do not need the launch angle... the only real purpose of the launch angle is to break up the launch velocity into horizontal and vertical velocity components... but they've already done that for you :)

Answer 7

vert velocity = 20 m/s. That's initial vertical velocity, final when it hits the ground will be equal but opposite, -20m/s

\[\Large v_f=v_i+at\] vf = -20m/s, vi=20m/s and a = -10m/s. Solve for t.


Then use \[\Large \Delta x=v_x t\] where delta x is your distance from launch point to end point, vx = 30m/s (initial horizontal velocity)

Answer 8

t should be 4 seconds - an object fired at 20m/s vertically will take 2 seconds to reach it's max height, and another 2 seconds to reach the ground again (time to ascend is equal to time to descend)