A small ball is thrown vertically upwards with an initial velocity of 40m/s. Find:(1) its velocity after moving 3 seconds(2) the maximum heigh attained(3) the time to attain the maximum height(4)the total time taken for the ball to return to the ground again
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To solve these problems, we'll use the basic equations of motion for an object in free fall.
1. Velocity after 3 seconds:
The equation we can use is:
Vf = Vi + at
where Vf is the final velocity after 3 seconds, Vi is the initial velocity, a is the acceleration, and t is the time.
Given:
Initial velocity, Vi = 40 m/s
Acceleration, a = -9.8 m/s^2 (negative because the object is moving against gravity)
Time, t = 3 s
Using the equation, we can calculate the final velocity:
Vf = Vi + at
Vf = 40 m/s - 9.8 m/s^2 * 3 s
Vf = 40 m/s - 29.4 m/s
Vf = 10.6 m/s
Therefore, after 3 seconds, the velocity of the ball is 10.6 m/s upwards.
2. Maximum height attained:
To find the maximum height, we can use the following equation:
h = Vi * t + 0.5 * a * t^2
where h is the maximum height, Vi is the initial velocity, a is the acceleration, and t is the time taken to reach the maximum height.
Given:
Initial velocity, Vi = 40 m/s
Acceleration, a = -9.8 m/s^2
Time to reach maximum height, t = ?
At the maximum height, the velocity is zero (since the ball momentarily stops before falling down).
Using the above equation, we can solve for t:
0 = 40 m/s * t + 0.5 * (-9.8 m/s^2) * t^2
This equation is a quadratic equation, which upon solving will give us the value of t. Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 0.5 * (-9.8 m/s^2), b = 40 m/s, and c = 0.
Solving the quadratic equation, we find two possible solutions for t. However, since the time cannot be negative, we only consider the positive solution.
(t = 40 m/s + √(40 m/s)^2 - 4 * 0.5 * (-9.8 m/s^2) * 0) / (2 * 0.5 * (-9.8 m/s^2))
t = 40 m/s + √(1600 m^2/s^2) / (-9.8 m/s^2)
t = 40 m/s + 40 m/s / (-9.8 m/s^2)
t = 40 m/s - 4.08 s
t ≈ 35.92 s
Therefore, the time taken to reach the maximum height is approximately 35.92 seconds.
Now, substitute this value of t back into the equation for height to find the maximum height:
h = 40 m/s * (35.92 s) + 0.5 * (-9.8 m/s^2) * (35.92 s)^2
h ≈ 716.8 m
Therefore, the maximum height attained by the ball is approximately 716.8 meters.
3. Time to attain the maximum height:
As solved above, the time to reach the maximum height is approximately 35.92 seconds.
4. Total time taken to return to the ground:
To find the total time taken for the ball to return to the ground, we can simply double the time taken to reach the maximum height.
Total time = 2 * time to reach maximum height
Total time = 2 * 35.92 s
Total time ≈ 71.84 s
Therefore, the total time taken for the ball to return to the ground again is approximately 71.84 seconds.