Projectile motion locust jump question

AI Thread Summary
To determine the initial velocity of a locust that can jump 75 cm at a 55° angle, the equations of projectile motion were applied. The calculations involved setting the time equations equal to each other and cross-multiplying to isolate the initial velocity. The resulting value was found to be approximately 4.0 m/s, but the textbook solution indicated a different initial velocity of 2.8 m/s using the range formula R = v²(sin²Ɵ) / 2g. This discrepancy highlights the importance of correctly applying the formulas for projectile motion. Understanding these calculations is essential for solving similar physics problems effectively.
The_big_dill
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Homework Statement


A Locust can jump as far as 75 cm horizontally, given a 55° angle above horizontal, what is the initial velocity?

Given:
d=75cm (0.75m)

Angle=55°

Homework Equations


a= v/t

d=vt

The Attempt at a Solution



t = (v)sin55 / 9.81

t = 0.75 / (v)cos55

t = t

(v)sin55 / 9.81 = 0.75 / (v)cos55

(cross multiply)

((v)sin55) ((v)cos55) = (9.81) (0.75)

v²(0.47) = 7.36

v = √ (7.36 / 0.47)

v = 4.0m/s
 
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Figured it out :)

The textbook (McGraw-Hill Physics 12) only introduced the formula which makes it VERY easy to solve this question in the next few pages.

Solution:

R = v²(sin²Ɵ) / 2 (g)

(Plug in given values and solve for unknown variable)

v = 2.8 m/s
 

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