A tennis ball of mass 100g is dropped from a height of 40m.It hits the ground and bounces back to a height of 30m after an impact of 0,01s.Ignore the air resistance .what was the velocity of the ball when it hit the ground ?
You have more information than you need to answer this question. The time to fall does not depend upon the mass, or how high it bounces, or the contact time. The "fall time" is simply
t = sqrt(2H/g)= 2.86 seconds
Are there other parts of the question that require the other information (such as the average force on the ground during a bounce)?
22.98
To find the velocity of the ball when it hit the ground, we can use the principles of physics.
First, let's calculate the time it takes for the ball to hit the ground. We can use the equation:
\[ h = \frac{1}{2}gt^2 \]
where \( h \) is the height, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( t \) is the time.
Substituting the given values, we get:
\[ 40 = \frac{1}{2} \times 9.8 \times t^2 \]
Simplifying the equation, we have:
\[ t^2 = \frac{40 \times 2}{9.8} \]
\[ t^2 = \frac{80}{9.8} \]
\[ t^2 \approx 8.16 \]
\[ t \approx \sqrt{8.16} \]
\[ t \approx 2.86 \text{ s} \]
Next, we can find the velocity with which the ball hit the ground using the equation:
\[ v = gt \]
where \( v \) is the velocity and \( t \) is the time.
Substituting the values, we have:
\[ v = 9.8 \times 2.86 \]
\[ v \approx 28.05 \text{ m/s} \]
Therefore, the velocity of the ball when it hit the ground is approximately 28.05 m/s.