A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 9.50 m/s at an angle of 24.0° below the horizontal. It strikes the ground 5.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
43.39 m
(b) Find the height from which the ball was thrown.
141.945 m
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?
My answers for parts A and B are correct; I can't figure out C.
I tried using the equations delta y= Voy*t + (1/2)g*t^2, using 0 and 9.5sin24 as Voy, but neither worked. I solved the quadratic and got two irrational solutions or times greater than 5 seconds. My last attempt I used t=(2*distance/g)^(1/2)= 1.427seconds; this sounds somwhat logical. Could this be correct?
For (c), did you use -9.8 m/s^2 as the coefficient of t^2? . Did you use -10 m for y? You must consider the upward initial velocity component. Using the quadratic equation, I get a positive root of t=1.87 s.
I did try using both positive and negative g and 10. I also did one positive with one negative; I tried everything. This assignment is online, so I plugged in your answer and that's incorret also. I don't understand, but thanks for the help.
To calculate the time it takes for the ball to reach a point 10.0 m below the level of launching, we can use the kinematic equation for vertical motion:
y = voy * t + (1/2) * a * t^2
In this equation, y represents the displacement in the vertical direction (which is -10.0 m since we are measuring downwards), voy is the initial velocity in the y direction (which is -9.5 m/s * sin(24°)), t is the time, and a is the acceleration in the y direction (which is -9.8 m/s^2).
Plugging in the values, the equation becomes:
-10.0 = -9.5 * sin(24°) * t + (1/2) * (-9.8) * t^2
Simplifying the equation, we get:
4.9 * t^2 - 2.412 * t - 10 = 0
Now, we can solve this quadratic equation for t.
Using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a
where a = 4.9, b = -2.412, and c = -10, we can calculate the values to find the roots of the equation.
The positive root will give us the time it takes for the ball to reach the point 10.0 m below the level of launching.
Substituting the values, we get:
t = (-(-2.412) ± √((-2.412)^2 - 4 * 4.9 * -10)) / (2 * 4.9)
t = (2.412 ± √(5.825344 + 196)) / 9.8
t = (2.412 ± √(201.825344)) / 9.8
t = (2.412 ± 14.200) / 9.8
Now, we calculate the two possible solutions for t:
t1 = (2.412 + 14.200) / 9.8
t1 = 16.612 / 9.8
t1 ≈ 1.69 s
t2 = (2.412 - 14.200) / 9.8
t2 = -11.788 / 9.8
t2 ≈ -1.20 s
Since we are considering the time it takes for the ball to reach a point below the level of launching, we discard the negative value for time. Therefore, the ball takes approximately 1.69 seconds to reach a point 10.0 m below the level of launching.