Answer:
(a) t = 3.74 s
(b) H = 136.86 m
(c) Vₓ = 41.83 m/s, Vy = 0 m/s
(d) ax = 0 m/s², ay = 9.8 m/s²
Explanation:
(a)
Time to reach maximum height by the projectile is given as:
t = V₀ Sinθ/g
where,
V₀ = Launching Speed = 55.6 m/s
Angle with Horizontal = θ = 41.2°
g = 9.8 m/s²
Therefore,
t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)
<u>t = 3.74 s</u>
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(b)
Maximum height reached by projectile is:
H = V₀² Sin²θ/g
H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)
<u>H = 136.86 m</u>
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(c)
Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:
Vₓ = V₀ₓ = V₀ Cos θ
Vₓ = (55.6 m/s)(Cos 41.2°)
<u>Vₓ = 41.83 m/s</u>
Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.
<u>Vy = 0 m/s</u>
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(d)
Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.
<u>ax = 0 m/s²</u>
The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:
<u>ay = 9.8 m/s²</u>
