A ball is thrown straight up at 24.0 m/s. Ignore the effects of air resistance in your calculations.After 2.00 s?
after 2 seconds, what? Height?
hf=hi+vi*t-4.8t^2
= 0+24*2-4.8*4
To find the height of the ball after 2.00 seconds, we can use the equation of motion for a vertically thrown object. The equation is given by:
h = h0 + v0*t + (1/2)*g*t^2
Where:
- h is the height of the ball at time t
- h0 is the initial height (which we'll assume is 0 m since it is thrown straight up)
- v0 is the initial velocity of the ball (in this case, 24.0 m/s)
- t is the time (2.00 s)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Now, let's plug in the values into the equation:
h = 0 + (24.0 m/s)*(2.00 s) + (1/2)*(9.8 m/s^2)*(2.00 s)^2
First, calculate the term with the initial velocity:
h = 0 + (24.0 m/s)*(2.00 s) + (1/2)*(9.8 m/s^2)*(4.00 s^2)
h = 0 + 48.0 m + (1/2)*(9.8 m/s^2)*(16.0 s^2)
Next, calculate the term with the acceleration due to gravity:
h = 48.0 m + (1/2)*(9.8 m/s^2)*(16.0 s^2)
h = 48.0 m + (1/2)*(9.8 m/s^2)*(256.0 s^2)
h = 48.0 m + 1,254.4 m
Finally, add up the two terms to find the height of the ball after 2.00 seconds:
h = 1,302.4 m
Therefore, the height of the ball after 2.00 seconds is approximately 1,302.4 meters.