Question

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?

Answer 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.

Answer 2

youre wrong. the cliff is 4.66m tall

Answer 3

How?

Answer 4

How

Answer 5

give this question answer

Answer 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.

Answer 7

Tite