Playing jackstone as a kid, you discovered that there is a certain time before the rubber ball falls and you took advantage of this when picking up the stars on the floor. If you threw the ball 4.92m/s up into the air, how fast will it be moving at a height of 53.86cm? How much time has elapsed upon reaching this height?
v = Vi + a t = 4.92 - 9.81 t
h = 0 + Vi t - 4.9 t^2 so 0.538 = 4.92 t - 4.9 t^2
4.9 t^2 - 4.92 t + 0.538 = 0
solve quadratic for t
then calculate v
https://www.mathsisfun.com/quadratic-equation-solver.html
roots; time = .125 and .879 seconds
use .125 because .879 is on the way down
so v = 4.92 - 9.81 (.125)
why did gravity become 4.9?
To find out how fast the ball will be moving at a height of 53.86 cm, we can use the principles of projectile motion and conservation of energy.
1. Calculate the initial velocity (v₀) of the ball when it was thrown upwards. In this case, the initial velocity is 4.92 m/s.
2. Determine the height at which you want to know the velocity. In this case, it is 53.86 cm, which is equivalent to 0.5386 m.
3. Use the concept of conservation of energy. At the highest point of the ball's trajectory, all of its initial kinetic energy is converted into gravitational potential energy. Thus, we can write the equation as:
v₁² = v₀² - 2gh,
where v₁ is the final velocity at the specified height, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
4. Substitute the known values into the equation:
v₁² = (4.92 m/s)² - 2 × 9.8 m/s² × 0.5386 m.
5. Solve for v₁²:
v₁² = 24.2064 m²/s² - 10.53724 m²/s².
v₁² = 13.66916 m²/s².
6. Take the square root of v₁² to get the final velocity, v₁:
v₁ = √13.66916 m²/s².
v₁ ≈ 3.70 m/s.
Therefore, the ball will be moving at approximately 3.70 m/s at a height of 53.86 cm.
To calculate the time elapsed upon reaching this height, we can use the kinematic equation:
v = v₀ - gt,
where v is the final velocity (which we just calculated as 3.70 m/s), g is the acceleration due to gravity (9.8 m/s²), v₀ is the initial velocity (4.92 m/s), and t is the time.
1. Rearrange the equation to solve for t:
t = (v₀ - v) / g.
2. Substitute the known values into the equation:
t = (4.92 m/s - 3.70 m/s) / 9.8 m/s².
3. Calculate t:
t = 1.22 m/s / 9.8 m/s².
t ≈ 0.1245 s.
Therefore, the time elapsed upon reaching a height of 53.86 cm is approximately 0.1245 seconds.