A projectile is fired with a speed of 50 m/s at an angle of 37

Question

find the horizontal and vertical components of its velocity

find the time that the projectile will spend in the air

find the distance downfield that it will travel

Answer 1

horizontal (until it hits the ground) U = 50 cos 37 = 39.9 m/s

vertical at t = 0 , Vo = 50 sin 37 = 30.1 m/s

vertical problem
v = Vo - 9.8 t
at top, v = 0
0 = 30.1 - 9.8 t
so
t = 3.07 seconds at top
add another 3.07 seconds to fall
total time in air = 6.14 seconds

horizontal problem
travels at 39.9 m/s for 6.14 seconds so goes
39.9*6.14 = 245 meters

Answer 2

Thanks

Answer 3

sorry but i don't get it
can show the equations u used
without the values or explain it in words or something

Answer 4

hmm, well
draw a right triangle with hypotenuse 50 at 37 degrees up from horizontal
then sin 37 = Vo/50
and cos 37 = U/50
where Vo is initial speed up
and U is horizontal speed, period (no horizontal forces so no change in horizontal speed.)

Now in the vertical direction
v = Vo - gt = Vo - 9.8 t
from the general form
v = Vo + a t where a here is the acceleration of gravity, g, 9.8 m/s^2 and is down, so negative

Now when the thing gets as high as it goes, the vertical speed is zero (it stops moving up)
so
0 = Vo - 9.8 t at the top.

It takes it just as long to fall as to rise, so double that time to get total time in the air.
If you do not belive that work it all out
z = Zo + Vo t + (1/2)a t^2
z = 0 at start and 0 at finish so
0 = 0 + Vo t - (9.8/2) t^2
or
t = 0 and
t = Vo/(9.8/2) = = 2 Vo/9.8
or twice Vo/9.8 as we did before

then once you have the time in the air multiply horizontal speed by time to get horizontal distance

Answer 5

thanks