Question

Please help mee.

A fisherman casts a lure toward the river at an angle of 25° above the horizontal. The lure reaches a maximum height of 2.9 m above where it was cast.

What is the initial velocity of the lure? Assume that the line exerts no appreciable drag force on the hook.​

Answer 1

17.84 m/s.

Step-by-step explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 25°

Maximum height (H) reached = 2.9 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =..?

Thus, we can obtain the initial velocity of the lure as follow:

H = u² sine² θ /2g

2.9 = u² × (sine 25)² / 2×9.8

2.9 = u² × (0.4226)²/19.6

Cross multiply

u² × (0.4226)² = 2.9 × 19.6

u² × (0.4226)² = 56.84

Divide both side by (0.4226)²

u² = 56.84/(0.4226)²

u² = 318.27

Take the square root of both side

u = √318.27

u = 17.84 m/s.

Therefore, the initial velocity of the lure is 17.84 m/s.