A ball thrown with the speed of 50m/s attained the height of 100m. Calculate the time of flight
To calculate the time of flight of a ball thrown with a speed of 50 m/s and attaining a height of 100 m, we can use the kinematic equation for vertical motion.
The equation is given as:
h = ut + (1/2)gt^2
Where:
h = height (in this case, 100 m)
u = initial velocity (in this case, 50 m/s)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time of flight (which we want to find)
Since the ball is thrown upwards, the initial velocity (u) and acceleration due to gravity (g) are in opposite directions. Therefore, we need to take the signs into account.
Substituting the values into the equation, we get:
100 = (50)t - (1/2)(9.8)t^2
Rearranging the equation to form a quadratic equation:
0 = (1/2)(9.8)t^2 - (50)t + 100
This equation can be solved using either factoring, completing the square, or using the quadratic formula. However, since factoring might be more complex in this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where a = (1/2)(9.8), b = -50, and c = 100.
Plugging in the values, we get:
t = (-(-50) ± √((-50)^2 - 4(1/2)(9.8)(100))) / (2(1/2)(9.8))
Simplifying further:
t = (50 ± √(2500 - 980)) / (9.8)
t = (50 ± √(1520)) / (9.8)
Calculating the square root, we get:
t = (50 ± 38.987) / 9.8
Therefore, the time of flight of the ball is either:
t ≈ (50 + 38.987) / 9.8 ≈ 8.77 seconds
or
t ≈ (50 - 38.987) / 9.8 ≈ 1.27 seconds
So, the time of flight of the ball is approximately 8.77 seconds or 1.27 seconds.