Trained dolphins are capable of a vertical leap of 7.0m straight up from the surface of the water-an impressive feat. suppose you could train a dolphin to launch itself out of the water at this same speed but at an angle. what maximum horizontal range could the dolphin achieve?

   θ = 45º

Explanation:

To find the solution, let's use the projectile launch equation

    R = vo² sin 2θ / g

Where vo is the initial speed of the dolphin, T is the angle of the jump and g the gravity acceleration.

To obtain a maximum range, the sine T function must be 1,

     sin 2θ = 1

     2θ = sin⁻¹ 1

     2θ = 90

     θ = 45º

Therefore, the dolphin should jump at an angle of 45º from the horizontal

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

a displacement distance is size of displ