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A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.2 $\mathrm{m}$ have been recorded. If a frog jumps 2.2 $\mathrm{m}$ and the launch angle is $45^{\circ},$ find $(\mathrm{a})$ the frog's launch speed and (b) the time the frog spends in the air. Ignore air resistance.

(a) $v _ { 0 } = 9.29 \mathrm { ms } ^ { - 1 }$

(b) $t = 1.34 \mathrm { s }$

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the party the horizontal range of a projectile, knowing that Delta wise equaling zero meters, essentially it is fired at the same level at which it lands. And so we can say that Then Delta X is gonna be equaling the initial squared sine of tooth Ada, divided by G. And essentially this becomes the initial equaling the square root of G Delta X. This would be divided by sign of to theatre. And so we can then say that the initial it's gonna be equal in the square root of 9.80 meters per second squared multiplied by 2.2 meters. This would be divided by sign of two multiplied by 40 five degrees. And this is giving us 4.6 meters per second. This would be our final answer for part A. So so for the time of flight, we know that then, for part B, we can say doubt acts is equaling the initial co sign of data Delta T or simply tea. And so the find t This would be equaling Delta X divided by the initial co sign of 45 degrees. And so this is gonna be equaling. We could say 2.2 meters and then this would be divided bye. 4.6 meters per second multiplied by co sign of 45 degrees. And this is giving us 0.67 seconds. This would be our final answer for part B. That is the end of the solution. Thank you for watching.