Answer:
A) The maximum height reached by the projectile is 34.3 m.
B) The total time in the air is 5.29 s.
C) The range of the projectile is 144 m.
D) The speed of the projectile 1.80 s after firing is 28.4 m/s.
Explanation:
Please, see the attached figure for a better understanding of the problem.
The position and velocity vectors of the projectile at time "t" are as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t"
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = vector position at time t
Let´s place the origin of the frame of reference at the launching point so that x0 and y0 = 0.
A) At the maximum height, the vertical component of the velocity is 0 (see figure). Then, using the equation for the y-component of the velocity vector, we can obtain the time at which the projectile is at its maximum height:
vy = v0 · sin α + g · t
0 = 37.6 m/s · sin 43.6° - 9.8 m/s² · t
- 37.6 m/s · sin 43.6° / -9.8 m/s² = t
t = 2.65 s
The height of the projectile at this time will be the maximum height. Then, using the equation of the y-component of the vector position:
y = y0 + v0 · t · sin α + 1/2 · g · t² (y0 = 0)
y = 37.6 m/s · 2.65 s · sin 43.6° - 1/2 · 9.8 m/s² · (2.65)²
y = 34.3 m
The maximum height reached by the projectile is 34.3 m.
B) When the projectile reaches the ground, the y-component of the position vector is 0 (see vector "r final" in the figure). Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 = 37.6 m/s · t · sin 43.6° - 1/2 · 9.8 m/s² · t²
0 = t · (37.6 m/s · sin 43.6° - 1/2 · 9.8 m/s² · t) (t = 0, the initial point)
0 = 37.6 m/s · sin 43.6° - 1/2 · 9.8 m/s² · t
- 37.6 m/s · sin 43.6° /- 1/2 · 9.8 m/s² = t
t = 5.29 s
The total time in the air is 5.29 s.
C) Having the total time in the air, we can calculate the x-component of the vector "r final" (see figure) to obtain the horizontal distance traveled by the projectile:
x = x0 + v0 · t · cos α
x = 0 m + 37.6 m/s · 5.29 s · cos 43.6°
x = 144 m
The range of the projectile is 144 m.
D) Let´s find the velocity vector at that time:
v = (v0 · cos α, v0 · sin α + g · t)
vx = v0 · cos α
vx = 37.6 m/s · cos 43.6°
vx = 27.2 m/s
vy = v0 · sin α + g · t
vy = 37.6 m/s · sin 43.6° - 9.8 m/s² · 1.80 s
vy = 8.29 m/s
Then, the vector velocity at t = 1.80 s will be:
v = (27.2 m/s, 8.29 m/s)
The speed is the magnitude of the velocity vector:
[tex]|v| =\sqrt{(27.2 m/s)^{2} +(8.29 m/s)^{2}} = 28.4 m/s[/tex]
The speed of the projectile 1.80 s after firing is 28.4 m/s.
