Question - Calculating Acceleration Due to Gravity from Velocity and Time
Solution:
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.
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