Question

DUE IN 3 HOURS! PLEASE HELP AND SOLVE A-E.

For a projectile with initial velocity of 51.5 m/s and angle of elevation equal to 4 degrees.
a) Compute the horizontal velocity of the projectile in m/s.
b) Compute the vertical velocity in m/s.
c) Compute the time of flight for the projectile.
d) Compute the distance downrange.
e) Compute the maximum height reached by the projectile.

Answer 1

Below

Step-by-step explanation:

c) To find the time of flight of a projectile (assuming from the given information that there is no vertical displacement) you can use this formula :

t = 2(Vi) Sinx / g

t = 2(51.5m/s) Sin(4) / 9.8 m/s^2

t = 0.733 seconds

t = 0.73 seconds

d) To find the displacement in the x-direction you can use this formula :

dx = vi^2 sin(2x) / g

dx = (51.5m/s)^2 sin(2(4)) / 9.8m/s^2

dx = 37.6654 m

dx = 37.7 meters

e) To find the peak height of the projectile you can use this formula :

v^2sin^2(x) = 2gh

(51.5m/s)^2sin^2(4) = 2(9.8m/s^2)h

h = 0.658 meters

h = 0.66 meters

Hope this helps! Best of luck <3