3. In the given picture you see the motion path of cannonball. Find the maximum height it can reach, horizontal distance it covers and total time from the given information. The angle between cannonball and horizontal is 53°.

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Chapter1: Units, Trigonometry. And Vectors
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Please use the formulas in the second picture. 

Please write known and unknown like y = ?, x = ?, etc. too and show work in an easy to understand way.

3.
In the given picture you see the motion path of cannonball. Find the
maximum height it can reach, horizontal distance it covers and total time from
the given information. The angle between cannonball and horizontal is 53°.
V=25m/s
ttotal?
hmax=?
h=60m
X=?
Transcribed Image Text:3. In the given picture you see the motion path of cannonball. Find the maximum height it can reach, horizontal distance it covers and total time from the given information. The angle between cannonball and horizontal is 53°. V=25m/s ttotal? hmax=? h=60m X=?
Projectile Motion, Constant Accel. Formulas
X components (horizontal):
projectile:
Vx =Vx0 + a,t
X = X, + Vxot + ½ a,t?
v,? = V,0² + 2a,(x-x)
Vy =Vx0
X
X = X, + Vxot
v,? = v,
%D
2
x0
Y components (vertical):
projectile:
Vy = Vyo + a,t
y = yo + Vyot + ½ a,t²
v, = v,² + 2a,(y-y.)
Vy = Vyo - gt
y = yo + Vyot - ½ gt?
v, = v,² -
2g(y-yo)
= V.
= V.
Transcribed Image Text:Projectile Motion, Constant Accel. Formulas X components (horizontal): projectile: Vx =Vx0 + a,t X = X, + Vxot + ½ a,t? v,? = V,0² + 2a,(x-x) Vy =Vx0 X X = X, + Vxot v,? = v, %D 2 x0 Y components (vertical): projectile: Vy = Vyo + a,t y = yo + Vyot + ½ a,t² v, = v,² + 2a,(y-y.) Vy = Vyo - gt y = yo + Vyot - ½ gt? v, = v,² - 2g(y-yo) = V. = V.
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