Find how long does it take to reach.

Question

Your friend throws a ball straight up in the air with an initial speed of 67.0m/s. How long does it take for the ball to be 2.00m above the ground?

Answer 1

h(t) = -4.9t^2 + 67t

-4.9t^2 + 67t - 2 = 0

about t= 2/67 = 1/33 = 0.03 seconds, not long. after all at 67 m/sec to go 2 m, even with gravity makes gravity's deceleration nearly zero. -4.9(.2/67)^2 + 67(2/67) -2 = -4.9(2/67)^2 = - 0.004 which is close enough to zero.

or use the quadratic formula

t= 67/9.8 + (1/9.8)sqr(67^2 -4(4.9)2))

t = 6.84 + or - (1/9.8)sqr(4449.8) = 6.84 - 6.81 = 0.03 seconds

or if you want to round off to the nearest tenth or decimal place, it's t = 0.00. It's already so close to 2 meters, it's virtually already there.

Answer 2

Gravity g = 9.81 m/s²

Distance s(t) = -gt²/2 + v₀t

2 = -4.905t² + 67t

4.905t² -67t +2 = 0

t = 0.03, 13.6 seconds

t = 0.03 seconds is going up

t = 13.6 seconds is coming down

Answer 3

This question is a bit ambiguous, because it doesn't seem to take into account the height of the friend throwing the ball. Furthermore, it raises the question of how to set your coordinate system for the problem. For the sake of simplicity, let's assume that ground level is y = 0m and let's neglect the fact that if a human is throwing the ball, it is already higher than ground level.

Okay, then let's organize what we need to find and what we know about the problem:

V₀ = 67 m/s is the initial speed of the ball

y = 2m is the height of the ball at the time we are looking for.

a = -g = -9.8 m/s² since we assigning up as the positive direction.

Now what formulas do we know that have those known quantities?

There are two that look like promising candidates;

1. In the absence of other forces, y = V₀t
2. In general though, y = V₀t + 1/2 a t²

Using equation 2, we can plug our values into each variable and solve for t using the quadratic equation.

This leads to t @ 2m being t = 0.30s

Answer 4

y = 67t - 4.9t² = 2, so we have equation for time : 4.9t² - 67t + 2 = 0, t = 0.03 s or t = 13.644 s