A ball is launched straight up from the edge of a cliff that is 250 m high. The ball's motion is given by s = 35t - 4.9t^2 where s is the ball's height in meters above the cliff top at time t seconds.
With what velocity does the ball strike the ground at the base of the cliff?
Answer: -285 m/sec
Is that correct?
Nope. That's way too fast.
How long does it take to fall to a height of -250m?
35t - 4.9t^2 = -250
t = 11.56
Now,
v = 35 - 9.8t = -78.26
To find the velocity with which the ball strikes the ground at the base of the cliff, we can first find the time it takes for the ball to reach the ground. We know that the height of the ball at any given time is given by the equation s = 35t - 4.9t^2.
When the ball reaches the ground, its height will be 0. So we can set the equation equal to 0 and solve for t:
0 = 35t - 4.9t^2
To solve this quadratic equation, we can either factoring or use the quadratic formula. Factoring might not be possible or easy in this case, so let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our equation, a = -4.9, b = 35, and c = 0. Plugging these values into the formula:
t = (-35 ± sqrt(35^2 - 4 * -4.9 * 0)) / (2 * -4.9)
Simplifying further, we have:
t = (-35 ± sqrt(1225)) / (-9.8)
Since we're looking for the time it takes for the ball to reach the ground, we can ignore the negative value:
t = (-35 + sqrt(1225)) / (-9.8)
t = (-35 + 35) / (-9.8)
t = 0 / (-9.8)
t = 0
This implies that the ball takes 0 seconds to reach the ground, which doesn't make sense in this context. It seems there may be an error in the problem statement or equation provided since a ball cannot reach the ground in zero seconds from a non-zero height.
Therefore, the answer provided (-285 m/s) is not correct.