A projectile is fired with an initial speed of 48.6 m/sat an angle of 44.2 ∘ above the horizontal on a long flat firing range.
Part A
Determine the maximum height reached by the projectile.


ymax = 
nothing 
m 
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Part B
Determine the total time in the air.


t = 
nothing 
s 
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Part C
Determine the total horizontal distance covered (that is, the range).


Δx = 
nothing 
m 
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Part D
Determine the speed of the projectile 1.30 s after firing.


v = 
nothing 
m/s 
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Part E
Determine the direction of the motion of the projectile 1.30 s after firing.


θ = 
nothing 
∘ above the horizontal 
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here,
the initial speed , u= 48.6 m/s
theta = 44.2 degree
a)
the maximum height reached by the projectile , hm = (u * sin(theta))^2 /( 2g)
hm = ( 48.6 * sin(44.2))^2 /(2*9.81)
hm = 58.5 m
b)
the total time in air , t = (2 * u * sin(theta))/g
t = 2 * 48.6 * sin(44.2)/9.81
t = 6.91 s
c)
the total horizontal distance covered , R = u * cos(theta) * t
R = 48.6 * cos(44) * 6.91 m = 241.6 m
d)
t1 = 1.3 s
the vertical speed , vy = u * sin(theta)  g * t1
vy = 48.6 * sin(44.2)  9.81 * 1.3 m/s = 21.13 m/s
the horizontal speed , vx = u * cos(theta) = 34.96 m/s
the speed of projectile , v = sqrt(vx^2 + vy^2)
v = sqrt(34.96^2 + 21.13^2) = 40.85 m/s
e)
the direction of motion , theta = arctan(21.13 /34.96) = 31.1 degree above the horizontal
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