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mariarad [96]
3 years ago
12

Wiley Coyote has missed the elusive road runner once again. This time, he leaves the edge of the cliff at 52.4 m/s with a purely

horizontal initial trajectory. If the canyon is 141 m deep, how far from the base of the cliff does the coyote land
Physics
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

<h2>280.86 m</h2>

Explanation:

Range is defined as the distance covered in the horizontal direction. In projectile, range is expressed as x = vt where;

x is the range

v is the velocity of the runner

t is the time taken

Before we can get the range though, we need to find the time taken t using the relationship S = ut + 1/2gt²

if u = 0

S = 1/2gt²

2S = gt²

t² = 2S/g

t = √2S/g

t = √2(141)/9.8

t = √282/9.8

t = 5.36secs

The range x = 52.4*5.36

x =280.86 m

Hence, the coyort lies approximately 280.86 m from the base of the cliff

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