a stone is thrown vertically upward,went up 98m and came down.How long was it in air?
To find out how long the stone was in the air, we need to use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
Where:
- h is the total height the stone traveled (98m)
- u is the initial velocity of the stone when it was thrown upwards (unknown)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time the stone was in the air (unknown)
Since the stone is thrown upwards, its final velocity will be 0 when it reaches its highest point. Therefore, we can use this information to find the initial velocity (u).
At the highest point, the final velocity (vf) can be found using the equation:
vf = u + gt
Since the final velocity (vf) is 0 at the highest point, we can use this information to solve for the initial velocity (u):
0 = u + (9.8 m/s^2) * t_max
Simplifying the equation, we find:
u = -9.8 m/s^2 * t_max
Now, let's calculate the time it took for the stone to reach 98m above its starting point.
Using the equation h = ut + (1/2)gt^2, we substitute the given values:
98m = u * t_max + (1/2) * (-9.8 m/s^2) * t_max^2
Since we know the value of u in terms of t_max, we can substitute it into the equation:
98m = (-9.8 m/s^2 * t_max) * t_max + (1/2) * (-9.8 m/s^2) * t_max^2
This equation is a quadratic equation in terms of t_max. Let's simplify it and solve for t_max:
98m = -9.8 m/s^2 * t_max^2 + (-4.9 m/s^2) * t_max^2
Combine like terms:
0 = -14.7 m/s^2 * t_max^2
Divide both sides by -14.7 m/s^2:
0 = t_max^2
Taking the square root of both sides, we get:
t_max = 0s
Here, we obtained t_max = 0s, which indicates that the stone took no time to reach its highest point and came back down instantly. However, this result seems contradictory to reality.
This contradiction arises because we ignored the air resistance while calculating. In real-world scenarios, air resistance plays a crucial role and affects the time of flight of the stone. Thus, considering air resistance will lead to a more accurate result.
I suspect it was twice as long as it took to fall 98m.
time to fall 98m
98=1/2 g t^2
t= sqrt(2*98/9.8)
so double that, and you have it.