Two children are playing on two trampolines. The first child can bounce 1.6 times higher than the second child. The initial speed up of the second child is 5.2 m/s.
(a) Find the maximum height the second child reaches.
(b) What is the initial speed of the first child?
(c) How long was the first child in the air?
To solve these questions, we can use the formulas for projectile motion. Let's break down each question step-by-step:
(a) Find the maximum height the second child reaches.
To find the maximum height, we can use the formula:
max height = (initial velocity^2 * sin^2(angle theta)) / (2 * acceleration due to gravity)
Given:
Initial speed of the second child (u2) = 5.2 m/s
The angle theta is not given, but we know that the first child can bounce 1.6 times higher than the second child. This means that the angle of projection for the first child will be the same as the second child, but its initial velocity will be higher.
Using the given information and the equation above, we can plug in the values:
max height (h2) = (5.2^2 * sin^2(theta)) / (2 * acceleration due to gravity)
(b) What is the initial speed of the first child?
Since the initial speed of the first child is 1.6 times higher than the initial speed of the second child, we can calculate it by multiplying the initial speed of the second child by 1.6:
initial speed of the first child (u1) = 1.6 * 5.2
(c) How long was the first child in the air?
To calculate the time of flight, we can use the formula:
time of flight = (2 * initial velocity * sin(angle theta)) / acceleration due to gravity
Using the given values, we can plug in the values:
time of flight = (2 * initial speed of the first child * sin(theta)) / acceleration due to gravity
Now, let's find the value of each part step-by-step.
To find the answers to the given questions, we can use the principles of projectile motion and apply the equations of motion. Let's solve each question step by step:
(a) Find the maximum height the second child reaches:
The maximum height reached by an object in projectile motion can be found using the formula:
H = (V^2 * sin^2(θ)) / (2 * g)
where H is the maximum height, V is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
In this case, since we are considering vertical motion, the launch angle is 90 degrees, and the sine of 90 degrees is 1. So the formula simplifies to:
H = (V^2) / (2 * g)
Given:
Initial speed of the second child, V = 5.2 m/s
Acceleration due to gravity, g = 9.8 m/s^2
Substituting these values into the formula:
H = (5.2^2) / (2 * 9.8)
H = 2.704 m
Therefore, the maximum height reached by the second child is approximately 2.704 meters.
(b) What is the initial speed of the first child:
It is given that the first child can bounce 1.6 times higher than the second child. Since both children are on trampolines, their initial velocities can be assumed to be equal. Let's denote the initial speed of the first child as V1.
Using the given information, we can set up the following equation:
V1 = 1.6 * V
where V is the initial speed of the second child.
Substituting the given value, V = 5.2 m/s, into the equation, we can solve for V1:
V1 = 1.6 * 5.2
V1 = 8.32 m/s
Therefore, the initial speed of the first child is approximately 8.32 m/s.
(c) How long was the first child in the air:
To find the time of flight for the first child, we can use the formula:
t = 2 * V * sin(θ) / g
Since the launch angle for vertical motion is 90 degrees, the sine of 90 degrees is 1. The formula thus simplifies to:
t = 2 * V / g
Substituting the given value, V = 8.32 m/s, into the formula:
t = 2 * 8.32 / 9.8
t = 1.70 seconds
Therefore, the first child was in the air for approximately 1.70 seconds.
Yo= 0m
h=?
A=g=9.8m/s^2
Child A:
Voa= ?
Ya= 1.6h
Ta= ?
Child B:
Vob= 5.2 m/s
Yb= h
a) Vob^2 = 2AYb
Yb= Vob/2A
h= (5.2 m/s)^2/(2(9.8m/s^2))
h= 1.37m
b) Voa=?
Voa^2=2AYa
Voa= sqrt(2AYa)
Voa= sqrt(2(9.8m/s^2)(1.6(1.37)))
Voa= 6.55m/s
c) sorry but i can't figure out how to do c