Question

Bull a is drop form a balcony 19,6 m from the ground at the same time and identical b is projected vertical downwards from the top of a building a 29, 6 m from the ground

Answer 1

To solve this problem, we need to consider the motion of both objects and determine when they will meet.

Let's start by considering the motion of bull A. It is dropped from a balcony 19.6 m high. We can use the equation of motion for free fall to determine the time it takes for bull A to reach the ground.

Using the equation: h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

19.6 = (1/2)(9.8)t^2
t^2 = 2(19.6)/9.8
t^2 = 4
t = 2 seconds

Therefore, bull A will take 2 seconds to reach the ground.

Now let's consider the motion of bull B. It is projected vertically downwards from the top of a building 29.6 m high. Since bull B is projected downwards, its initial velocity (u) will be negative.

We can use the equation: h = ut + (1/2)gt^2

29.6 = -ut + (1/2)(9.8)t^2

To find the time of flight for bull B, we need to rearrange the equation to solve for t:

(1/2)(9.8)t^2 - ut + 29.6 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac))/(2a)

where a = 1/2, b = -u (negative initial velocity), and c = 29.6.

Plugging in the values:

t = (-(-u) ± √((-u)^2 - 4(1/2)(29.6)))/(2(1/2))

t = (u ± √(u^2 - 59.2))/1

Since we know bull B is projected downwards, its initial velocity (u) will be negative.

Let's assume the initial velocity of bull B is -v (positive value). Plugging in the value, we get:

t = (-(-v) ± √((-v)^2 - 59.2))/1
t = (v ± √(v^2 - 59.2))/1

Now, we know that bull B will take the same amount of time as bull A to reach the ground, which is 2 seconds.

Therefore, we can set up the equation:

2 = (v ± √(v^2 - 59.2))/1

Solving this equation will give us the values of v, which represents the speed at which bull B is projected downwards from the building.