Asked by jess
                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!
            
        Help me please!
Answers
                    Answered by
            Anonymous
            
    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
    
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
                    Answered by
            Anonymous
            
    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
    
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
                    Answered by
            Anonymous
            
    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
    
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
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