If a ball is thrown straight up at a speed of 27.8 m/s, how long (in seconds) will it continue to rise before it starts to fall back down
To determine how long the ball will continue to rise before it starts to fall back down, we can use the kinematic equation for vertical motion. The equation we'll use is:
v = u + at
Where:
v = final velocity (0 m/s, as the ball starts to fall)
u = initial velocity (27.8 m/s, as the ball is thrown straight up)
a = acceleration due to gravity (-9.8 m/s², as gravity will act in the downward direction)
t = time in seconds (what we're trying to find)
Plugging in the values, we can solve for t:
0 m/s = 27.8 m/s + (-9.8 m/s²) * t
Rearranging the equation:
-9.8 m/s² * t = -27.8 m/s
Dividing both sides by -9.8 m/s²:
t = (-27.8 m/s) / (-9.8 m/s²)
t ≈ 2.84 seconds
Therefore, the ball will continue to rise for approximately 2.84 seconds before it starts to fall back down.
To find the time it takes for the ball to reach its highest point and start falling back down, you can use the equation of motion for vertical motion:
v = u + at
where:
v = final velocity (at the highest point, the velocity is 0)
u = initial velocity (27.8 m/s)
a = acceleration due to gravity (-9.8 m/s^2, assuming that the ball is thrown on Earth)
t = time
Rearranging the equation:
0 = 27.8 - 9.8t
Now, we can solve for 't'.
First, move 27.8 to the other side of the equation:
9.8t = 27.8
Next, divide both sides of the equation by 9.8:
t = 27.8 / 9.8
Using a calculator, the answer is approximately:
t ≈ 2.837 seconds
Therefore, the ball will continue to rise for approximately 2.837 seconds before it starts to fall back down.