(a) The velocity and position function of the pellet at a time , t ≥0, is (30, 30, 80) + 5t and (30, 30, 80)t + 5t² respectively.
(b) The time of flight of the pellet is 0.78 s and the range is 23.4 m.
(c) The maximum height reached by the pellet is 45.92 m.
Time of motion of the pellet
The time of motion of the pellet is calculated as follows;
t = √(2h/g)
t = √(2 x 3 / 9.8)
t = 0.78 second
Velocity of the pellet after 0.78 second
v = u + at
v = (30, 30, 80) + 5t
v = (30, 30, 80) + 5(0.78)
v = (30, 30, 80) + 3.9
v = (33.9, 33.9, 83.9) m/s
Position of the pellet
x = vt
x = [(30, 30, 80) + 5t]t
x = (30, 30, 80)t + 5t²
x = (33.9 x 0.78, 33.9 x 0.78, 83.9 x 0.78)
x = (26.44, 26.44, 65.44) m
Range of the projectile
R = Vₓt
where;
- Vₓ is horizontal velocity
- t is time of motion
R = 30 m/s x 0.78 s
R = 23.4 m
Maximum height of pellet
v² = u² - 2gh
where;
- v is the final velocity at maximum height = 0
- u is the initial vertical velocity = 30 m/s
- h is the maximum height
- g is acceleration due to gravity
0 = u² - 2gh
2gh = u²
h = u²/2g
h = (30²) / (2 x 9.8)
h = 45.92 m
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