A ball is thrown directly downward with an initial speed of 8.95 m/s, from a height of 29.0 m. After what time interval does it strike the ground?
use 2nd equation of motion.
S=UT+1/2ATSQURED
h = Hi + Vi t - 4.9 t^2
0 = 29 - 8.95 t - 4.9 t^2
t = [8.95+/-sqrt(8.95^2 + 4*29*4.9)]/(2*4.9)
t = [8.95 +/-25.46 ] /9.81
use + t
3.51 seconds
the answer is not 3.51 seconds when i type in the answer it doesn't work
it's a quadratic
9.8 t^2 + 8.95 t -29 = 0
Answer works out to 1.32 sec
To find the time interval after which the ball strikes the ground, we can use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
Where:
- h is the initial height (29.0 m)
- u is the initial velocity (8.95 m/s, in the downward direction)
- g is the acceleration due to gravity (-9.8 m/s^2, in the downward direction)
- t is the time interval we want to find
We can rearrange the equation to solve for t:
(1/2)gt^2 + ut - h = 0
This is a quadratic equation, which can be solved using the quadratic formula:
t = (-u ± sqrt(u^2 - 4gh)) / (2g)
Now, we can substitute the given values into the equation and calculate the time interval.
t = (-8.95 ± sqrt((8.95)^2 - 4 * (-9.8) * (-29.0))) / (2 * -9.8)
t = (-8.95 ± sqrt(79.8 + 1139.2)) / -19.6
t = (-8.95 ± sqrt(1219)) / -19.6
Since we are looking for the time interval for the ball to strike the ground, we take the positive root:
t = (-8.95 + sqrt(1219)) / -19.6
Calculating this expression gives us:
t ≈ 1.84 seconds
Therefore, the ball will strike the ground approximately 1.84 seconds after it is thrown downward.