A grasshopper leaps into the air from the edge of a vertical cliff, as shown in the figure below (Figure 1). Assume that d1 = 7.26 cm, d2 = 1.11 m, and that θ = 48 ∘

Question

1.) what is the initial speed of the grasshopper

2.) what would the height of the cliff be

Answer 1

Assuming that d1 is the vertical distance of the max y from the starting point and d2 is the horizontal distance from the starting point after jumping:

1) What is the initial speed of the grasshopper

v_y(t) = v_o,y + t*a(t); For projectile motion, a(t) = -g = -9.81 m/s²

To find v_{o,y ,}

vo,y = √2g(y - y_o) = √2g(7.26 * 0.01 - 0) = 1.1935 m/s

Now use θ to find v_o,x to calculate v_o

tan(48º) = 1.1935/v_o,x

v_o,x = 1.0746 m/s

v_o = √(1.1935)² + (1.0746)²

v_o = 1.6060 m/s

2) What would the height of the cliff be?

When x(t) = d2, y(t) = 0

0 = y_o + (1.1935)t - .5gt²

Solve for t using x(t)

x(t) = d2 = x_o + (1.0746)t, x_o = 0

d2 = 1.11 = (1.0746)t --> t = 1.032 s

Plug in 1.032 for t in y(t) and solve for y_o

0=y_o+(1.1935)(1.032)−12(9.81)(1.032)²

y_o = -((1.1935)(1.032)−12(9.81)(1.032)²)



y_o = 3.992 m