A ball is thrown from the top of a building with an initial velocity of 30m/s straight upward, at an initial height of 50m above the ground. The ball just misses the edge of the roof on is way down and hits the ground. Determine: a) time needed for ball to reach max height. B)the max height. C) time needed for ball to hit ground. D) velocity and diplacement ( magnitude and direction) of the ball at t=sec( measured from beginning. Please explain step by step
To solve this problem, we can use the kinematic equations of motion.
a) Time needed for ball to reach max height:
The initial velocity (u) of the ball is 30 m/s, and the acceleration due to gravity (a) is -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity). The initial position (s) is 50 m.
We will use the equation:
v = u + at
Since the ball is coming to a stop at its maximum height, its final velocity (v) will be 0 m/s. Rearranging the equation, we get:
t = (v - u) / a
Substituting the given values, we have:
t = (0 - 30) / -9.8
Calculating this, we find:
t = 3.06 seconds
So, it takes 3.06 seconds for the ball to reach its maximum height.
b) The maximum height:
To find the maximum height (h), we can use the equation:
s = ut + (1/2)at^2
At the maximum height, the final velocity is 0 m/s. So, the formula becomes:
s = ut + (1/2)at^2
0 = 30(3.06) + (1/2)(-9.8)(3.06)^2
Solving this equation, we find:
s = 46.53 meters
Therefore, the maximum height reached by the ball is 46.53 meters.
c) Time needed for ball to hit the ground:
To find the time it takes for the ball to hit the ground, we can use the equation:
s = ut + (1/2)at^2
In this case, the initial position is 50 meters (above the ground) and we want to find the time it takes for the ball to reach the ground, so the final position (s) will be 0 meters. The acceleration due to gravity (a) is -9.8 m/s^2.
0 = 30t + (1/2)(-9.8)t^2
Simplifying this equation, we have:
4.9t^2 - 30t = 0
Factoring out t, we get:
t(4.9t - 30) = 0
From this equation, we can see that there are two possible solutions: t = 0 or t = 6.12 seconds. However, t = 0 represents the initial time when the ball was thrown, so we discard that solution. Therefore, the time needed for the ball to hit the ground is 6.12 seconds.
d) Velocity and displacement of the ball at t=2 sec:
To find the velocity (v) and displacement (s) at t = 2 seconds, we can use the equations:
v = u + at
s = ut + (1/2)at^2
Given:
u = 30 m/s
a = -9.8 m/s^2
t = 2 seconds
Calculating the formulas, we find:
v = 30 + (-9.8)(2)
v = 30 - 19.6
v = 10.4 m/s
s = 30(2) + (1/2)(-9.8)(2)^2
s = 60 + (1/2)(-9.8)(4)
s = 60 + (-19.6)
s = 40.4 meters
Therefore, at t = 2 seconds, the velocity of the ball is 10.4 m/s (upward) and the displacement is 40.4 meters (above the ground).
To determine the time needed for the ball to reach its maximum height, you can use the equation for vertical displacement (dy) of an object thrown upwards:
dy = viy * t - (1/2) * g * t^2
where
- dy is the vertical displacement,
- viy is the initial vertical velocity (positive upwards in this case),
- g is the acceleration due to gravity (approximately -9.8 m/s^2),
- t is the time.
Since the ball is thrown straight upward, its initial vertical velocity (viy) is 30 m/s. At the maximum height, the vertical velocity becomes zero (as the ball reaches its highest point and starts to fall). Therefore, we can set viy = 0 in the equation.
0 = 0 * t - (1/2) * g * t^2
0 = (-1/2) * g * t^2
To find the time needed for the ball to reach its maximum height, solve for t:
(-1/2) * g * t^2 = 0
t^2 = 0
Since the coefficient is zero, the value of t does not affect this equation. Therefore, the time needed for the ball to reach its maximum height is t = 0 seconds.
Now, let's move on to determining the maximum height (h) reached by the ball. You can use the equation for vertical displacement (dy) again, this time with viy = 30 m/s and t = 0 since we're calculating the displacement at the maximum height.
dy = viy * t - (1/2) * g * t^2
h = 30 * 0 - (1/2) * (-9.8) * 0^2
Since both terms in the equation are multiplied by zero, the calculation becomes:
h = 0 - 0
h = 0
Therefore, the maximum height reached by the ball is 0 meters. This means that the ball just misses the edge of the roof as it starts to descend.
Moving on to determining the time needed for the ball to hit the ground, we can use the equation for vertical displacement (dy), this time with the starting height of the ball (dy) as 50 m and final height at the ground (0 m).
0 = 30 * t - (1/2) * (-9.8) * t^2
Rearrange the equation to solve for t:
0 = 30t + 4.9t^2
0 = t(30 + 4.9t)
Applying the zero product property, we get:
t = 0 or t = -30/4.9
Since time cannot be negative in this context (as it represents the duration taken by the ball to reach the ground), we discard t = -30/4.9.
Therefore, t = 0 or t ≈ 3.06 seconds.
To determine the velocity of the ball at t = 1 second, we can use the equation for velocity (vy) at any given time:
vy = viy - g * t
Plug in the values:
vy = 30 - 9.8 * 1
vy = 30 - 9.8
vy ≈ 20.2 m/s (rounded to one decimal place)
Since the ball is thrown upwards, the velocity at t = 1 s is positive, so the velocity of the ball at this time is approximately 20.2 m/s in the upward direction.
For the displacement at t = 1 second, we can use the equation for displacement (dy) at any given time:
dy = viy * t - (1/2) * g * t^2
Plug in the values:
dy = 30 * 1 - (1/2) * 9.8 * 1^2
dy = 30 - 9.8 * 1
dy = 30 - 9.8
dy ≈ 20.2 m (rounded to one decimal place)
Since the displacement is positive, the ball has moved approximately 20.2 meters above its starting position at t = 1 second.