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Spittingcobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a spitting cobra rears up to a height of 0.500 $\mathrm{m}$ above the ground and launches venom at 3.50 $\mathrm{m} / \mathrm{s}$ , directed $50.0^{\circ}$ above the horizon. Neglecting air resistance, find the horizontal distance traveled by the venom before it hits the ground.

$d = 1.59 \mathrm { m }$

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So we'll first find the initial X and why components of the velocity. This would be equaling essentially the initial velocity co sign if ADA and for the wide component the initial velocity sign of data. And so this is gonna be equaling 3.50 meters per second again for both of them. And then this will be co sign of 50 degrees sign of 50 degrees for the X component. We then have 2.25 meters per second and for the wide component, we then have 2.68 meters per second. Now we can find the time of flight. Um, however, we need to first find the final velocity in the direction the final velocity in the UAE direction because we don't have time is gonna be equaling weaken. Say negative because we know that it's gonna be traveling downwards. This would be a negative square root of V. Why initial squared plus two times the acceleration in the why Direction times, delta y. And so from here we can then, of course, substitute. This would be equaling negative square root and this would be 2.68 meters per second quantity squared plus two multiplied by negative 9.80 meters per second squared multiplied by negative 0.500 meters and we have a final. Why velocity equaling negative 4.12 meters per second. Now we can solve. For the time of flight t, this would be equaling essentially v Y final minus the why initial divided by G and this would be equaling or better yet divided by the acceleration, the war direction. And so this would be equaling negative 4.12 meters per second minus 2.68 meters per second, divided by negative 9.80 meters per second squared and we find that t the time of flight is equaling 0.694 seconds. And so the horizontal distance traveled. During that time, Delta X would simply be equaling the ex initial times t, this is gonna be equaling 2.25 meters per second multiplied by 0.694 seconds and me find that Then the horizontal distance traveled Delta axes equaling 1.56 meters. This would be our final answer. That is the end of the solution. Thank you for watching