A ball is thrown vertically upwards in the air at a velocity of 50(m/s) and reaches a maximum height of 20(m). Calculate the time when the ball reaches the 10(m) mark, while going up and coming down. Use the following equation of motion:
y(f) = y(i) + v(i)t+1/2at^2
y(f) and y(i) = final and initial position, v(i) = initial velocity, a = acceleration, and t = time, respectively.
velocity is 50-9.8t.
v=0 at top of trajectory, at t=50/9.8=5.10 sec.
y = yi + 50t - 4.9t^2, so at t=5.10,
20 = yi + 50(5.10) - 4.9(5.10)^2
yi = -107.55
eh? thrown from below ground?
Yet it makes sense, since it was thrown upward with a velocity of 50 m.s, yet only made it 20m high. No mention was made of a horrific air resistance.
Anyway, assuming the above,
y = -107.55 + 50t - 4.9t^2
solve for t when y=10:
t = 3.67 or 6.53
If I botched it, make the fix.
To calculate the time when the ball reaches the 10m mark, you can use the equation of motion provided:
y(f) = y(i) + v(i)t + 1/2at^2
In this case, the initial position (y(i)) is 0m, as the ball starts from the ground. The initial velocity (v(i)) is 50m/s, and we need to calculate the time it takes for the ball to reach a position of 10m.
So, we have:
10 = 0 + 50t + 1/2at^2
Since the ball is going upwards initially, the acceleration (a) is equal to the acceleration due to gravity, which is approximately -9.8m/s^2 (negative because it acts in the opposite direction to the motion).
10 = 50t - 4.9t^2
To solve this quadratic equation, we can rearrange it to standard form:
4.9t^2 - 50t + 10 = 0
Now we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values:
a = 4.9
b = -50
c = 10
t = (50 ± √((-50)^2 - 4(4.9)(10))) / (2(4.9))
Simplifying further:
t = (50 ± √(2500 - 196)) / 9.8
t = (50 ± √(2304)) / 9.8
t = (50 ± 48) / 9.8
This gives two solutions for t:
t = 98 / 9.8 = 10s (when ball is going up)
t = 2 / 9.8 ≈ 0.20s (when ball is coming down)
Hence, the ball reaches the 10m mark while going up at 10 seconds and while coming down at 0.20 seconds.