Question

A Chinook salmon can jump out of water with a speed of 6.50 m/s . How far horizontally can a Chinook salmon travel through the air if it leaves the water with an initial angle of =30.0° with respect to the horizontal? (Let the horizontal direction the fish travels be in the + direction, and let the upward vertical direction be the + direction. Neglect any effects due to air resistance.)

Answer 1

Answer:

3.73m

Explanation:

What we are asked to find is the range covered by the fish. This is given by the following equation (1). Range can simply be defined as the horizontal distance covered by a body whose motion freely under gravity is in two dimensions. The motion of the fish is in two dimensions, the vertical dimension and the horizontal dimension.

$R=\frac{u^2sin2\theta}{g}.........(1)$

where u = 6.5m/s is the initial velocity, g is acceleration due to gravity which is taken as $9.8m/s^2$ and $\theta=30.0^o$.

Substituting these values into equation (1), we obtain the following;

$R=\frac{6.5^2sin2(3)}{9.8}\\R=\frac{42.25sin60}{9.8}\\\\R=\frac{36.59}{9.8}\\R=3.73m$

Answer 2

The range should be 3.73m.

Important information:

A Chinook salmon can jump out of water with a speed of 6.50 m/s . The initial angle of =30.0°

The range refers to the horizontal distance i.e. covered by a body whose motion freely under gravity should be in two dimensions. It can be the vertical dimension and the horizontal dimension.

calculation of the range:

$= 6.5sin^2\div 9.8\\\\= 36.59 \div 9.8$

= 3.73m

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