A projectile is fired with a speed of 50 m/s at an angle of 37
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
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
Thanks
sorry but i don't get it
can show the equations u used
without the values or explain it in words or something
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
thanks
To find the horizontal and vertical components of the velocity of the projectile, we can use the given speed and angle.
Horizontal component of velocity (Vx):
Vx = V * cos(theta)
Vertical component of velocity (Vy):
Vy = V * sin(theta)
where V is the initial speed of the projectile and theta is the angle of projection.
Given:
V = 50 m/s
theta = 37 degrees
1. Calculate the horizontal component of velocity (Vx):
Vx = V * cos(theta)
Vx = 50 * cos(37)
Vx ≈ 50 * 0.7986
Vx ≈ 39.93 m/s
2. Calculate the vertical component of velocity (Vy):
Vy = V * sin(theta)
Vy = 50 * sin(37)
Vy ≈ 50 * 0.6018
Vy ≈ 30.09 m/s
Now, let's calculate the time the projectile will spend in the air.
3. The time of flight (T) can be found using the formula:
T = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
T = (2 * Vy) / g
T = (2 * 30.09) / 9.8
T ≈ 6.14 seconds
Finally, let's calculate the horizontal distance traveled by the projectile (or range).
4. The horizontal distance (d) can be obtained using the formula:
d = Vx * T
d = 39.93 * 6.14
d ≈ 245.39 meters
Therefore, the horizontal component of velocity is approximately 39.93 m/s, the vertical component of velocity is approximately 30.09 m/s, the time of flight is approximately 6.14 seconds, and the horizontal distance traveled is approximately 245.39 meters.