An archerfish, peering from just below the water surface, sees a grasshopper standing on a tree branch that's just above the water. The archerfish spits a drop of water at the grasshopper and knocks it into the water. The grasshopper's initial position is 0.45 m above the water surface and 0.25 m horizontally away from the fish's mouth. The launch angle of the drop is 63° with respect to the water surface. Anticipating that the grasshopper would do this, the archerfish decides to spit the water at a higher speed but at the same launch angle (63°). At what speed should the archerfish spit the water for the drop to hit the grasshopper?

Question

Physics

Paxton Hardy

17 October, 19:50

An archerfish, peering from just below the water surface, sees a grasshopper standing on a tree branch that's just above the water. The archerfish spits a drop of water at the grasshopper and knocks it into the water. The grasshopper's initial position is 0.45 m above the water surface and 0.25 m horizontally away from the fish's mouth. The launch angle of the drop is 63° with respect to the water surface. Anticipating that the grasshopper would do this, the archerfish decides to spit the water at a higher speed but at the same launch angle (63°). At what speed should the archerfish spit the water for the drop to hit the grasshopper?

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Answers (1)

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Karen Oconnor · Answer 1

let the velocity of throw is u.

Time to reach horizontally. 25 m is equal to reach. 45 m vertically

t =.25 / u cos63

=.55 / u

For vertical motion

- h = - ut + 1/2 gt²

-.45 = - usin63 x. 55 / u +.5 x9.8 (.55 / u) ²

-.45 = -.49 + 1.48 / u²

.04 = 1.48 / u²

u² = 37

u = 6.08 m / s