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
(1) Vo - gt = 40 - 9.8*3 = 10.6 m/s
The direction will still be upwards, since the number is positive.
(2) The height increase H will be such that added potential energy
M*g*H equals initial kinetic energy, (M/2*(Vo^2).
H = Vo^2/(2g) = 81.6 m
(3) Maximum height occurs when V = 0
Vo - gt = 0
t = 4.08 s.
(4)Time going up = Time coming down.
Total time = 2*4.08 = 8.16 s
To solve this problem, we can use the equations of motion for vertical motion.
1) To find the velocity after 3 seconds, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the ball is moving vertically upwards, the acceleration due to gravity is -9.8 m/s^2.
Substituting the given values:
v = 40 m/s - 9.8 m/s^2 * 3 s
v = 40 m/s - 29.4 m/s
v = 10.6 m/s
Therefore, the velocity after 3 seconds is 10.6 m/s upwards.
2) To find the maximum height attained, we can use the equation:
s = ut + (1/2)at^2
where s is the displacement (height), u is the initial velocity, a is the acceleration, and t is the time. Since the ball is moving vertically upwards, the acceleration due to gravity is -9.8 m/s^2. The initial displacement is zero because the ball starts from the ground.
Substituting the given values:
s = 40 m/s * 3 s + (1/2) * (-9.8 m/s^2) * (3 s)^2
s = 120 m - 44.1 m
s = 75.9 m
Therefore, the maximum height attained is 75.9 meters.
3) The time taken to attain the maximum height can be found using the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. At the maximum height, the final velocity is zero.
Substituting the given values:
0 = 40 m/s + (-9.8 m/s^2) * t
t = 40 m/s / 9.8 m/s^2
t = 4.08 s
Therefore, it takes approximately 4.08 seconds for the ball to reach the maximum height.
4) The total time taken for the ball to return to the ground can be found using the equation:
s = ut + (1/2)at^2
Since the ball takes the same amount of time to reach its maximum height and to return to the ground, we can multiply the time to reach the maximum height by 2.
Total Time = 4.08 s * 2
Total Time = 8.16 s
Therefore, the total time taken for the ball to return to the ground is approximately 8.16 seconds.
To solve these problems, we can use the equations of motion for an object moving vertically with constant acceleration due to gravity.
First, let's consider the direction of motion. Since the ball is thrown vertically upwards, the acceleration due to gravity acts in the opposite direction, downwards. We can assume the positive direction as upwards and the acceleration due to gravity as negative.
Now, let's break down the problems step by step:
(1) To find the velocity after 3 seconds:
We have the initial velocity (u) as 40 m/s and the time (t) as 3 seconds.
The equation we can use is:
v = u + at,
where:
v = final velocity,
u = initial velocity,
a = acceleration,
t = time.
Since the ball is moving against gravity, the acceleration would be -9.8 m/s^2.
Plugging in the values, we get:
v = 40 m/s - (9.8 m/s^2) * 3 s = 40 m/s - 29.4 m/s = 10.6 m/s.
So, the velocity of the ball after 3 seconds is 10.6 m/s upwards.
(2) To find the maximum height attained:
To find the maximum height, we need to use the equation:
h = u*t + (1/2)*a*t^2,
where:
h = height,
u = initial velocity,
t = time,
a = acceleration.
At the highest point, the velocity will be zero (v = 0 m/s).
Plugging in the values into the equation, we have:
0 = 40 m/s * t + (1/2) * (-9.8 m/s^2) * t^2.
Simplifying the equation, we get:
-4.9 t^2 + 40t = 0.
Factorizing it, we have:
t * (10t - 40) = 0.
This gives us two solutions: t = 0 (initial time) and t = 4 seconds.
We discard t = 0 since it doesn't represent the height attained. Hence, the time to attain the maximum height is 4 seconds.
Then, we can use the equation:
h = u*t + (1/2)*a*t^2,
h = 40 m/s * 4 s + (1/2) * (-9.8 m/s^2) * (4 s)^2.
Calculating this equation, we get:
h = 80 m - 78.4 m = 1.6 m.
Therefore, the maximum height attained is 1.6 meters.
(3) To find the time taken to attain the maximum height:
We have already found this in the previous step. The time taken to attain the maximum height is 4 seconds.
(4) To find the total time taken for the ball to return to the ground again:
To find the total time taken, we need to consider both the upward and downward journey of the ball.
The time taken for the upward journey is the time to attain the maximum height, which we found to be 4 seconds.
The time taken for the downward journey can be found using the equation:
v = u + at,
where:
v = final velocity (0 m/s when the ball hits the ground),
u = initial velocity (velocity at the highest point),
a = acceleration (-9.8 m/s^2),
t = time.
Plugging in the values, we have:
0 = u - 9.8 m/s^2 * t.
Rearranging the equation, we get:
u = 9.8 m/s^2 * t.
From the previous step, we found that the maximum height is 1.6 meters. At the highest point, the velocity is zero.
Using the equation:
v^2 = u^2 + 2ah,
where:
v = final velocity,
u = initial velocity,
a = acceleration,
h = height,
we can find the initial velocity at the highest point:
0 = u^2 + 2 * (-9.8 m/s^2) * (1.6 m).
Simplifying the equation, we get:
u = sqrt(2 * 9.8 * 1.6) m/s = sqrt(31.36) m/s ≈ 5.6 m/s.
Now, we can calculate the time taken for the downward journey:
0 = 5.6 m/s - 9.8 m/s^2 * t.
Solving for t, we get:
t = 5.6 m/s / 9.8 m/s^2 ≈ 0.57 seconds.
Therefore, the total time taken for the ball to return to the ground is the sum of the time taken for the upward journey (4 seconds) and the time taken for the downward journey (0.57 seconds), which is approximately 4.57 seconds.