a rock is thrown vertically upwards with an initial velocity of 30 m/s. Neglecting air resistance, calculate
a) the maximum height reached
b) the time taken to reach maximum height
v1 = 30 m/s
v2 = 0 m/s
a = negative gravity
simple plug n chug to find height (distance) and time
To calculate the maximum height reached by the rock and the time taken to reach that height, we can use the following equations of motion:
a) Maximum height reached (distance):
The equation for calculating height is given by:
\[v^2 = u^2 + 2as\]
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (30 m/s)
a = acceleration (acceleration due to gravity, taken as -9.8 m/s^2 since it acts in the opposite direction)
s = distance or height
Substituting the given values into the equation:
\[0^2 = 30^2 + 2(-9.8)s\]
\[0 = 900 - 19.6s\]
\[19.6s = 900\]
\[s = \frac{900}{19.6} \approx 45.92 \, \text{m}\]
Therefore, the maximum height reached by the rock is approximately 45.92 meters.
b) Time taken to reach maximum height:
The equation for calculating time is given by:
\[v = u + at\]
Where:
v = final velocity (0 m/s at maximum height)
u = initial velocity (30 m/s)
a = acceleration (acceleration due to gravity, taken as -9.8 m/s^2 since it acts in the opposite direction)
t = time taken
Substituting the given values into the equation:
\[0 = 30 + (-9.8)t\]
\[-30 = -9.8t\]
\[t = \frac{-30}{-9.8} \approx 3.06 \, \text{s}\]
Therefore, the time taken to reach the maximum height is approximately 3.06 seconds.