owenleak owenleak

07-07-2020
Mathematics

contestada

What is the start velocity of a javelin accelerated at 3.75 m/s2 for 2 seconds, and reaching 9375 m/s?

Respuesta :

saraharif006 saraharif006

07-07-2020

Answer:

Step-by-step explanation:

u = ?

a = 3.75m/s^2

t = 2 s

v = 9375m/s

a = v-u/t

3.75 = 9375 - u / 2

3.75 × 2 = 9375 - u

7.5 = 9375 - u

u = 9375 - 7.5

= 9367.5 m/s

Hope this helps

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