A snowball is thrown from the ground into the air with a velocity of 20.0 m/s at an angle of 27.0 degrees to the horizontal. What is the maximum height reached by this object?
Help me please!
simpler answer.
we dont need the x component because we are looking at the max height which means we only look at the y component.
Vy = 20 sin 27 = 9.08 m/s
a = -9/8 m/s^2
Vmax = 0m/s (because this is the maximum point the object will reach and then fall back down to earth.)
use: V2^2 = V1^2 + 2ad
d = -(9.08)^2 / 2(-9.81m/s^2) ---> (rearranged)
d = 4.2m
Therfore the max height is 4.2m
All that matters is the vertical problem. The horizontal speed is constant until the crash.
sq
Initial vertical speed = Vi = 20 sin 27 = 9.08 m/s
v = Vi + a t
here on earth a is about -9.81 m/s^2
at the top v = 0
so
a t = - Vi
t = -9.08 / -9.81 of a second
then
h = 0 + Vi t+ (1/2) a t^2
or
h max = Vi t -4.9 t^2
alternately the average speed up = (Vi + 0) /2 = Vi/2
so
h max = (Vi/2) t
To find the maximum height reached by the snowball, you can use the equations of motion in both the vertical and horizontal directions.
First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component of the velocity can be calculated as:
Vx = V * cos(θ)
Vx = 20.0 m/s * cos(27.0°)
Vx ≈ 17.95 m/s (rounded to two decimal places)
The vertical component of the velocity can be calculated as:
Vy = V * sin(θ)
Vy = 20.0 m/s * sin(27.0°)
Vy ≈ 9.15 m/s (rounded to two decimal places)
Now, let's find the time it takes for the snowball to reach its maximum height using the vertical motion equation:
Vy = Vy0 + gt
0 m/s = 9.15 m/s - 9.8 m/s² * t
Solving for t:
9.8 m/s² * t = 9.15 m/s
t ≈ 0.9347 s (rounded to four decimal places)
Next, we can find the maximum height (H) using the vertical motion equation:
H = Vy0 * t + (1/2) * g * t²
H = 9.15 m/s * 0.9347 s + (1/2) * 9.8 m/s² * (0.9347 s)²
H ≈ 4.085 m (rounded to three decimal places)
Therefore, the maximum height reached by the snowball is approximately 4.085 meters.