If the ball was thrown with an initial velocity of 20m/s and an initial velocity horizontal velocity of 30m/s.what will be the total time of the flight of the ball? consider that the ball was released from the ground and landed some meters away.
To find the total time of flight for the ball, we can use the equation of motion for vertical motion.
The equation we can use is:
h = v*t + (1/2)*g*t^2
Where:
h = height reached by the ball (considering the ball was released from the ground and landed some meters away, h would be the maximum height reached by the ball)
v = initial vertical velocity (which is 20 m/s in this case)
t = time of flight (what we need to find)
g = acceleration due to gravity (which is approximately 9.8 m/s^2)
Since the initial vertical velocity is given, we know that the ball will reach its maximum height when it reaches the point where its vertical velocity becomes zero. At this point, the ball will start descending back to the ground.
At the maximum height, the vertical velocity will be zero. Therefore, we can write the equation as:
0 = v - g*t
Simplifying this equation, we get:
g*t = v
t = v/g
Now, we need to find the time taken for the ball to reach the maximum height and double that time to find the total time of flight.
Substituting the values into the equation:
t = 20 m/s / 9.8 m/s^2
t ≈ 2.04 seconds
Since the total time of flight is double the time taken to reach the maximum height:
Total time of flight ≈ 2 * 2.04 seconds
Total time of flight ≈ 4.08 seconds
Therefore, the total time of flight for the ball will be approximately 4.08 seconds.
I assume you meant an initial vertical velocity of 20 m/s. The horizontal velocity does not matter.
Since v = 30-9.81t
v=0 at t=3.06 seconds.
That is when it has reached the top of its arc.
So total flight time is twice that.