Student A is on a 9m high bridge while Student B is on the ground. Student B throws an apple up to Student B with a velocity of 18m/s. How long does it take for the apple to reach student A's hand? if student A is unable to catch the apple, how long would it take for the apple to come back to the ground?

See previous post.

To solve this problem, we need to analyze two scenarios separately:

Scenario 1: Calculating the time it takes for the apple to reach Student A's hand.
Scenario 2: Calculating the time it takes for the apple to come back to the ground.

Let's start with Scenario 1:

To calculate the time it takes for the apple to reach Student A's hand, we need to use the equation of motion:

h = ut + (1/2)gt^2

where:
h = height (distance) = 9m
u = initial velocity = 18m/s (being thrown upwards)
g = acceleration due to gravity = 9.8m/s^2
t = time taken

Since we are determining the time, we rearrange the equation as follows:

(1/2)gt^2 + ut - h = 0

Now we can substitute the values into the equation and solve for t using the quadratic formula:

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

Simplifying this equation further, we get:

t = (-u ± √(u^2 + 2gh)) / g

Substituting the values:

t = (-18 ± √(18^2 + 2 * 9.8 * 9)) / 9.8

After calculating this equation, we get two possible values for time: one positive and one negative. The negative value does not make sense in this context, so we take the positive value as the answer.

Now, let's move on to Scenario 2:

To calculate the time it takes for the apple to come back to the ground, we need to consider that Student B threw the apple upward with an initial velocity of 18m/s, and it will eventually fall back to the ground.

Using the equation of motion again:

h = ut + (1/2)gt^2

where:
h = height (distance) = 0m (ground level this time since the apple is coming back down)
u = initial velocity = 18m/s (this time downward due to gravity)
g = acceleration due to gravity = 9.8m/s^2
t = time taken

Simplifying the equation, we have:

(1/2)gt^2 + ut - h = 0

Substituting the values:

(1/2) * 9.8 * t^2 + (-18)t - 0 = 0

Solving this quadratic equation, we get two possible values for t: one positive and one negative. The negative value does not make sense in this context either, so we take the positive value as the answer.

By following these steps, we can calculate the time it takes for the apple to reach Student A's hand and the time it takes for the apple to come back down to the ground.