Playing jackstone as a kid, you discovered that there is a certain time before the rubber ball falls and you took advantage of this when picking up the stars on the floor. If you threw the ball (5.17)m/s up into the air, how fast will it be moving at a height of (61.70)cm? How much time has elapsed upon reaching this height?
see related questions below.
1. V^2 = Vo^2+2g*h = 5.17^2+(-19.6)0.617 = 14.64
V = 3.8 m/s.
2. V = Vo+g*T = 3.8
5.17-9.8T = 3.8
T =
To determine how fast the ball will be moving at a height of 61.70 cm, we can use the principles of projectile motion.
We know the initial velocity (u) of the ball when thrown upwards is 5.17 m/s. At its highest point, the velocity (v) will decrease to zero before it falls back down due to gravity.
To find the final velocity (v) at a specific height, we can use the following equation:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration due to gravity (approximately -9.8 m/s^2)
s = displacement
In this case, the displacement is the height of 61.70 cm, or 0.617 m. Rearranging the equation, we have:
v^2 = u^2 + 2as
v^2 = (5.17 m/s)^2 + 2(-9.8 m/s^2)(0.617 m)
v^2 = 26.7689 m^2/s^2 + (-12.0884 m^2/s^2)
v^2 = 14.6805 m^2/s^2
Taking the square root of both sides, we get:
v = √(14.6805 m^2/s^2)
v ≈ 3.8325 m/s
Therefore, the ball will be moving at approximately 3.8325 m/s when it reaches a height of 61.70 cm.
Now, let's calculate the time it takes for the ball to reach this height. We can use the following equation:
v = u + at
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (5.17 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
Substituting the known values, we have:
0 m/s = 5.17 m/s + (-9.8 m/s^2) * t
Rearranging the equation, we find:
-5.17 m/s = -9.8 m/s^2 * t
Dividing both sides by -9.8 m/s^2, we get:
t = -5.17 m/s / -9.8 m/s^2
t ≈ 0.5286 s
Therefore, approximately 0.5286 seconds have elapsed when the ball reaches a height of 61.70 cm.