You toss a tennis ball straight upward. At the moment it leaves your hand it is at a height of 1.5 m above the ground, and it is moving at a speed of 6.4 m/s.When the tennis ball is at a height of 2.2 m above the ground, what is its speed?
the ball's speed is being converted to gravitational potential energy
1/2 m 6.4^2 =
... m g (2.2 - 1.5) + 1/2 m v^2
6.4^2 = 1.4 g + v^2
v = Vi - 9.81 t
h = Hi + Vi t -4.9 t^2
====================
2.2 = 1.5 + 6.4 t - 4.9 t^2
solve quadratic for t
then
v = 6.4 - 9.81 t
To find the speed of the tennis ball when it is at a height of 2.2 m above the ground, we can use the principles of projectile motion.
First, let's establish the initial conditions:
- Height when the ball is thrown, h1 = 1.5 m above the ground
- Initial velocity when the ball is thrown, v1 = 6.4 m/s (upwards)
- Height when we want to find the speed, h2 = 2.2 m above the ground
Now, let's use the concept of conservation of mechanical energy. At any point during the ball's motion, the sum of its kinetic energy (KE) and potential energy (PE) remains constant:
PE1 + KE1 = PE2 + KE2
At the initial height (h1 = 1.5 m), the potential energy is given by the expression PE1 = m * g * h1, where m is the mass of the ball and g is the acceleration due to gravity (approximately 9.8 m/s^2).
At the highest point of the ball's motion (h2 = 2.2 m), the potential energy PE2 is given by PE2 = m * g * h2.
The kinetic energy at both points remains the same because the ball is moving at a constant speed throughout its motion.
Therefore, we can rewrite the equation as:
m * g * h1 + 0.5 * m * v1^2 = m * g * h2 + 0.5 * m * v2^2
Since mass (m) is common to both sides of the equation, we can cancel it out:
g * h1 + 0.5 * v1^2 = g * h2 + 0.5 * v2^2
Now, rearrange the equation to solve for v2:
0.5 * v2^2 = g * (h2 - h1) + 0.5 * v1^2
v2^2 = 2 * g * (h2 - h1) + v1^2
Finally, take the square root of both sides to get the value of v2:
v2 = √(2 * g * (h2 - h1) + v1^2)
Substituting the given values:
v2 = √(2 * 9.8 m/s^2 * (2.2 m - 1.5 m) + 6.4 m/s^2)
Simplifying the equation:
v2 = √(2 * 9.8 m/s^2 * 0.7 m + 6.4 m/s^2)
v2 = √(13.72 m^2/s^2 + 6.4 m^2/s^2)
v2 = √20.12 m^2/s^2
v2 ≈ 4.48 m/s
Therefore, when the tennis ball is at a height of 2.2 m above the ground, its speed is approximately 4.48 m/s.