a stone is projected at an angle of 60 degree and an initial velocity of 20 m/s .determine the time of flight
2×20×sin60=34.6410
Due to acceleration due to gravity
34.6410÷10=34.641s
the vertical speed is 20 sin 60° = 17.32 m/s
at time t, v = 17.32 - 9.8t
v=0 at t=1.767 s
That is when it has reached its maximum height.
Now, how long does it take to come back down?
TOF=2*20 sin60/10
Correct
Not too well understood
Plsssss explain to me
To determine the time of flight of a stone projected at an angle of 60 degrees and an initial velocity of 20 m/s, you can use the kinematic equations of motion.
The time of flight is the duration from when the stone is launched until it hits the ground again. In this case, we can split the motion into two components: horizontal and vertical.
First, let's find the time taken for the stone to reach its maximum height. At the maximum height, the vertical component of the velocity becomes zero.
We know that the initial velocity in the vertical direction (vy) is given by the equation:
vy = v * sin(θ)
where v is the initial velocity and θ is the angle.
Substituting the values, we have:
vy = 20 m/s * sin(60°)
vy = 20 m/s * 0.866
vy ≈ 17.3 m/s
The time taken to reach the maximum height can be found using the equation of motion:
vy = u + at
where u is the initial velocity in the vertical direction, a is the acceleration (which is equal to -9.8 m/s^2 due to gravity), and t is the time.
Substituting the values, we have:
0 = 17.3 m/s - 9.8 m/s^2 * t
Solving this equation for t, we have:
9.8 m/s^2 * t = 17.3 m/s
t ≈ 1.77 seconds
Next, we need to find the time taken for the stone to reach the ground from the maximum height. This is the same as the time taken to go up to the maximum height.
So the total time of flight (t_flight) is given by:
t_flight = 2 * t ≈ 2 * 1.77 seconds
t_flight ≈ 3.54 seconds
Therefore, the time of flight of the stone is approximately 3.54 seconds.