Projectile motion locust jump question

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SUMMARY

The initial velocity required for a locust to jump 75 cm horizontally at a 55° angle is calculated to be 4.0 m/s using the equations of projectile motion. The solution involves equating the time of flight derived from vertical motion to that from horizontal motion, leading to the equation v²(0.47) = 7.36. The final velocity can also be confirmed using the formula R = v²(sin²Ɵ) / 2g, resulting in a different value of 2.8 m/s, indicating a discrepancy that warrants further investigation.

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  • Knowledge of gravitational acceleration (9.81 m/s²)
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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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