Question

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!

Answer 1

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

Answer 2

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

Answer 3

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