components of the speed of the coin is given as
[tex]v_x = v cos60[/tex]
[tex]v_x = 6.4 cos60 = 3.2 m/s[/tex]
[tex]v_y = vsin60[/tex]
[tex]v_y = 6.4 sin60 = 5.54 m/s[/tex]
now the time taken by the coin to reach the plate is given by
[tex]t = \frac{\delta x}{v_x}[/tex]
[tex]t = \frac{2.1}{3.2}[/tex]
[tex]t = 0.656 s[/tex]
now in order to find the height
[tex]h = vy * t + \frac{1}{2} at^2[/tex]
[tex]h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2[/tex]
[tex]h = 1.52 m[/tex]
so it is placed at 1.52 m height
The height of the coin above the point where the coin leaves your hand is 1.53 meters
"We are to find the height of the coin above the point where the coin leaves your hand is 1.53 meters"
Velocity, V = 6.4m/s
Angle, θ = 60°
Resolve the coin speed into vertical and horizontal components :
Horizontal component (Vx) :
Vx = Vcosθ
Vx = 6.4cos60° = 6.4 × 0.5 = 3.20 m/s
Vertical component (Vy) :
Vy = Vsinθ
Vy = 6.4 × sin60° = 6.4 × 0.8660 = 5.54 m/s
Time taken by the coin to reach the plate :
Recall :
Speed = distance / time
Time taken = distance / speed
Horizontal Distance = 2.1 m
Speed = horizontal velocity = 3.2 m/s
Hence,
Time taken = 2.1 / 3.2 = 0.656 seconds
From the laws of motion :
S = ut + 0.5gt²
S = height or vertical distance
u = vertical speed
g = acceleration due to gravity = - 9.8m/s (downward motion)
S = (5.54 × 0.656) + (0.5 × - 9.8 × 0.656²)
S = 3.635625 - 2.11025390625
S = 1.52537109375
S = 1.53 meters
Hence, the height of the coin above the point where the coin leaves your hand is 1.53 meters.
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