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?

youre wrong. the cliff is 4.66m tall

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.

Tite

How?

How

give this question answer

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.