Question

An archer fish launches a droplet of water from the surface of a small lake at an angle of 60 above the horizontal. He is aiming at a juicy spider sitting on a leaf 46 cm to the east and on a branch 27 cm above the water surface. The fish is trying to knock the spider into the water so that the fish can eat the spider (a) What must the speed of the water droplet be for the the fish to be successful? m/s (b) When it hits the spider, is the droplet rising or falling? O rising falling

Answer 1

To find initial speed

In x-direction

Xf - Xo = Vx*t

given that Xf - Xo = 46 cm

Vx = V0*cos 60 deg = V0/2

0.46 = V0*t/2

V0*t = 0.92

In y-direction

Yf - Yo = Vy*t - 0.5*g*t^2

0.27 = V0*sin 60 deg*t - 0.5*9.8*t^2

0.27 = V0*t*0.866 - 4.9*t^2

from above equation V0*t = 0.92

0.27 = 0.92*0.866 - 4.9*t^2

t = sqrt [(0.92*0.866 - 0.27)/4.9] = 0.33 sec

So,

V0*t = 0.92

V0 = 0.92/0.33 = 2.79 m/sec