In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 m from this point. if you toss the coin with a velocity of 6.4 m> s at an angle of 60° above

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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components of the speed of the coin is given as

now the time taken by the coin to reach the plate is given by

now in order to find the height

so it is placed at 1.52 m height

0.55x745.7j and a calculator

it would be kevlar ^^