# Physics

A grasshopper leaps into the air from the edge of a cliff at a 50 degree angle. He reaches a maximum height 6.74 cm above the top of the cliff and travels a total horizontal distance of 1.06 m. How tall is the cliff and what is the initial speed of the grasshopper?

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1. Y^2 = Yo^2 + 2g*h = 0
Yo^2 - 19.6*0.0674 = 0
Yo^2 = 1.32
Yo = 1.15 m/s = Ver. component of initial velocity.

Yo = Vo*sin50 = 1.15
Vo = 1.15/sin50 = 1.50 m/s = Initial
velocity.

Xo = Vo*Cos50 = 1.50*Cos50 = 0.964 m/s =
Hor. component of initial velocity.

Dx = Xo*T = 1.06 m.
0.964*T = 1.06
T = 1.10 s. = Fall time.

h = 0.5g*T^2 - 0.0674 = 4.9*1.1^2 - 0.0674 = 5.60 m. = Ht. of the cliff.

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2. youre wrong. the cliff is 4.66m tall

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3. How?

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4. How

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6. Nathan is right, it's 4.66 meters tall. Since we are trying to find Y, and we also have all the different components of the kinematic already, we simply do Y= 0+1.15(1.1)+1/2(-9.8)(1.1)^2 which yields the result of (after removing the negative because we just want to know the height) 4.66m.

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