I have problems that i need solved as fast as possible and dont know what to do. please send answeres quick
A 57kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.7m/s. How fast is he going as he lands on the trampoline 3.8m below?
A 124g baseball is dropped from a tree 11.6m above the ground. With what speed would it hit the ground if air resistance could be ignored?
I just did the first on below.
second one
Again the mass does not matter
(1/2) m v^2 = m g h
v = sqrt (2 g h)
(remember that)
1. V^2 = Vo^2 + 2g*h = 0,
4.7^2 + (-19.6)h = 0,
h = 1.127 m. above the platform.
3.8 + 1.127 = 4.93 m. above gnd.,
V^2 = Vo^2 + 2g*h = 0 + 19.6*4.93 = 96.57,
V = 9.83 m/s.
2. V^2 = Vo^2 + 2g*h = 0 + 19.6*11.6 = ,
V = ?.
To solve these problems, we can use the laws of motion and basic formulas of kinematics. Let's tackle each problem step by step:
Problem 1:
We have the initial speed (4.7 m/s) and the vertical displacement (height difference between the platform and the trampoline, 3.8m). We need to find the final speed.
To solve this, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (what we want to find)
u = initial velocity (given as 4.7 m/s)
a = acceleration (due to gravity, approximately 9.8 m/s^2)
s = displacement (3.8 m)
Plugging in the values into the equation:
v^2 = (4.7 m/s)^2 + 2 * (-9.8 m/s^2) * (3.8 m)
Now, we can solve for v.
v^2 = 22.09 m^2/s^2 + (-74.84 m^2/s^2)
Adding the terms:
v^2 = -52.75 m^2/s^2
Taking the square root of both sides:
v = √(-52.75 m^2/s^2)
Here, we encounter a problem. The negative value under the square root indicates an imaginary number. It means that the trampoline artist does not reach the trampoline if we assume that air resistance is ignored. Therefore, we cannot calculate the final speed.
Problem 2:
We have the height from which the baseball is dropped (11.6m). We need to find the speed at which it hits the ground without considering air resistance.
To solve this, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (what we want to find)
u = initial velocity (0 since the ball is dropped)
a = acceleration (due to gravity, approximately 9.8 m/s^2)
s = displacement (11.6 m)
Plugging in the values into the equation:
v^2 = (0 m/s)^2 + 2 * 9.8 m/s^2 * (11.6 m)
Calculating the right side:
v^2 = 0 m^2/s^2 + 226.16 m^2/s^2
Simplifying:
v^2 = 226.16 m^2/s^2
Taking the square root of both sides:
v = √(226.16 m^2/s^2)
v = 15.05 m/s
Hence, if air resistance is ignored, the baseball would hit the ground with a speed of approximately 15.05 m/s.
Please note that in real-world scenarios, air resistance significantly affects the motion of objects, so these calculations are simplified versions that ignore its effects.