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
Answer:
D
Explanation:
I think the answer is D, because of hallucinations
V=r/t
Speed equals displacement over the time
V=100/9.92=10.08ms^-1
Answer:
The work done shall be 14715 Joules
Explanation:
The work done by a force 'F' in a displacement 'dy' is given by
At any position 'y' the weight shall be sum of weft of water and weight of string
Thus applying values we get
