In a cartoon program, Peter tosses his baby, Stewie, up into the air to keep the child entertained. Stewie reaches a maximum height of 0.787 m above the release point. Suppose the positive y axis points upward.
(a) With what initial velocity was Stewie thrown? (Express your answer in vector form.)
v = Vi - g t
h =Hi +Vi t - (1/2) g t^2
if g = 9.81 m/s^2
v = 0 at top
9.81 t = Vi so t = Vi / 9.81
Hi = 0
0.787 = Vi t - 4.9 t^2 = Vi^2 /9.81 - 4.9 Vi^2/9.81*2 = Vi^2/19.6
Vi = 3.93 m/s upward
or, if you want another formula, the max height
h = v^2 / 2g
0.787 = v^2/19.62
v^2 = 15.44
v = 3.93
you may already have seen this as relates to the usual horizontal mostion
v^2 = 2as
To find the initial velocity with which Stewie was thrown, we need to use the equations of motion. In this case, we can use the equation for the maximum height reached by an object thrown vertically upwards:
y_max = (v^2)/(2g)
Where:
- y_max is the maximum height reached
- v is the initial velocity of the object
- g is the acceleration due to gravity
In this problem, the maximum height reached by Stewie is given as 0.787 m. We can use the value of the acceleration due to gravity on Earth, which is approximately 9.8 m/s^2.
Plugging in the values, we have:
0.787 m = (v^2)/(2 * 9.8 m/s^2)
Simplifying the equation, we get:
v^2 = 0.787 m * 2 * 9.8 m/s^2
v^2 = 15.3628 m^2/s^2
Taking the square root of both sides, we find:
v = √(15.3628 m^2/s^2)
v ≈ 3.92 m/s
Therefore, Stewie was thrown upwards with an initial velocity of approximately 3.92 m/s in the positive y direction.