Calculating Acceleration Due to Gravity from Velocity and Time
The photo shows three questions, of which the first one (#4) is:
If a ball is dropped and attains a velocity of 29.4 m/s in 3.00 s, what is the acceleration due to gravity?
To solve this, you can use the following kinematic equation for uniformly accelerated motion:
\[ v = u + at \]
where:
- \( v \) is the final velocity
- \( u \) is the initial velocity (which is 0 m/s, since the ball is dropped)
- \( a \) is the acceleration (which we are trying to find, and we expect it to be the acceleration due to gravity)
- \( t \) is the time
Given that the initial velocity \( u = 0 \) m/s, the final velocity \( v = 29.4 \) m/s, and the time \( t = 3.00 \) s, you can plug these values into the equation to find \( a \):
\[ 29.4 = 0 + a \times 3.00 \]
Now solve for \( a \):
\[ a = \frac{29.4}{3.00} \]
\[ a \approx 9.8 \, \text{m/s}^2 \]
Thus, the acceleration due to gravity, \( g \), is approximately 9.8 m/s², which is in fact the standard value for gravitational acceleration near the Earth's surface.
