Question

A cat 0. 525 m from a base of the ledge jumps at a velocity of 7. 09 m/s at a 78. 5 degree angle. What is its height when it reaches the edge of the ledge.

Answer 1

The height reached by the cat when it reaches the edge of the ledge is 1.9 m.

The given parameters:

• Position of the cat, x = 0.525 m
• Velocity of the cat, v = 7.09 m/s
• Angle of projection, θ = 78.5⁰

Height attained by a projectile;

• This is the vertical displacement of the projectile from the point of projection.

The horizontal component of the velocity is calculated as follows;

$v_x = vcos (\theta)\\\\v_x = 7.09 \times cos(78.5)\\\\v_x = 1.41 \ m/s$

The time of motion of the projectile is calculated as follows;

$X = v_x t\\\\t = \frac{X}{v_x} \\\\t = \frac{0.525}{1.41} \\\\t = 0.37 \ s$

The height reached by the cat when it reaches the edge of the ledge is calculated as follows;

$h = v_yt - \frac{1}{2} gt^2 \\\\h = (7.09\times sin78.5 \times 0.37)- (\frac{1}{2} )(9.8)(0.37^2)\\\\h = 1.9 \ m$

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