Answer:
t = 0.11 second
The time taken to reach the maximum height of the trajectory is 0.11 second
Explanation:
Taking the vertical component of the initial velocity;
Vy = Vsin∅
Initial velocity V = 2.15 m/s
Angle ∅ = 30°
Vy = 2.15sin30 = 2.15 × 0.5
Vy = 1.075 m/s
The height of the rock at time t during the flight is;
From the equation of motion;
h(t) = Vy×t - 0.5gt^2
g= acceleration due to gravity = 9.8m/s^2
Substituting the given values;
h(t) = 1.075t - 0.5(9.8)t^2
h(t) = 1.075t - 4.9t^2
The rock is at maximum height when dh/dt = 0;
dh(t)/dt = 1.075 - 9.8t = 0
1.075 - 9.8t = 0
9.8t = 1.075
t = 1.075/9.8
t = 0.109693877551 s
t = 0.11 second
The time taken to reach the maximum height of the trajectory is 0.11 second
The time it takes for the rock to reach the maximum height is 0.11 seconds.
Projectile motion is the motion experienced by an object thrown into space and it moves along a curved path.
The time (T) for the rock to reach maximum height is given by:
[tex]T=\frac{usin(\theta)}{g}[/tex]
where u is the velocity = 2.15 m/s, g is the acceleration due to gravity = 9.81 m/s², θ = angle = 30 degree
[tex]T=\frac{usin(\theta)}{g}=T=\frac{2.15*sin(30)}{9.81}=0.11 \ s[/tex]
Hence the time it takes for the rock to reach the maximum height is 0.11 seconds.
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