A stone is thrown vertically upward with the velocity of 49.0 m/s. (a) how high will it travel before it starts to fall? (b) how many seconds will it take the stone to reach the highest point?
v = Vi - 9.81 t
at top v = 0
0 = 49 - 9.81 t
t = 49/9.81
h = 49 t - 4.9 t^2
To answer both questions, we can use the equations of motion for an object thrown vertically with constant acceleration due to gravity.
(a) To find how high the stone will travel before it starts to fall, we can use the formula for the maximum height reached by an object thrown vertically:
Δy = (v₀² - v²) / (2g)
Where:
Δy = Change in vertical position (maximum height) - this is what we are trying to find
v₀ = Initial velocity (49.0 m/s)
v = Final velocity (0 m/s at the highest point, when the stone starts falling)
g = Acceleration due to gravity (-9.8 m/s²)
Substituting the values into the formula, we get:
Δy = (49.0² - 0) / (2 * -9.8)
= (2401.0) / (-19.6)
≈ -122.5 meters
Note: The negative sign indicates that the displacement is in the opposite direction of the initial throw (upward).
Therefore, the stone will travel approximately 122.5 meters upwards before it starts to fall.
(b) To find how many seconds it will take for the stone to reach the highest point, we can use the formula for the time of flight of an object thrown vertically:
t = (v - v₀) / g
Where:
t = Time of flight (what we are trying to find)
v₀ = Initial velocity (49.0 m/s)
v = Final velocity (0 m/s at the highest point, when the stone starts falling)
g = Acceleration due to gravity (-9.8 m/s²)
Substituting the values into the formula, we get:
t = (0 - 49.0) / -9.8
= 49.0 / 9.8
≈ 5.0 seconds
Therefore, it will take approximately 5.0 seconds for the stone to reach the highest point.