Answer: a) vox = vo × cos θ, b) voy =vo× sin θ,
c) H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 m
Explanation:
A)
The velocity v0 is at angle θ to the horizontal.
The horizontal component of vo (vox), vo and the vertical component of vo (voy) all form a right angle triangle.
With vo as the hypotenus, vox as the adjacent and voy as the opposite.
To get vox, we relate vo and vox ( hypotenus and adjacent)
From trigonometry
Cos θ relates hypotenus and adjacent, hence we have that
Cos θ = vox/vo
vox = vo × cos θ
B)
To get the vertical component of vo, we relate vo and voy ( hypotenus and opposite).
According to trigonometry, sin θ relates hypotenus and opposite, hence we have that
Sin θ = voy/vo
voy =vo× sin θ
C)
The formulae for the maximum height of a projectile motion is given as
H = vo² (sin θ)²/2g
Where g = acceleration due to gravity = 9.8 m/s²
By substituting the parameters, we have that
H = 26² × (sin 17)²/2(9.8)
H = 676 × 0.0854/19.6
H = 57.7304/ 19.6
H = 2.94 m
D)
This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g
From the equation of motion
vy = voy - gt
0 = voy - gt
But voy = vo sinθ
0 = vo sinθ - gt
gt = vo sinθ
t = vo sinθ / g
E)
The horizontal distance covered formulae is given by
R = u² sin2θ/g
R = 26² × sin 2(17)/9.8
R = 676 × sin 34/ 9.8
R = 378.014/ 9.8
R = 38.57 m