<p>Q- A balloon is rising at 10m/s when its passenger throws a ball straight up at 12m/s .How much later does the passenger catch the ball? </p>
<p>thank you!!!</p>
<p>12 m/s relative to the balloon or relative to earth? makes a difference</p>
<p>wouldn't i need delta Y to get the time it reaches the peak?</p>
<p>I copied the question exactly. I think it is relative to the ballon. </p>
<p>I used delta x = 0 = v_initial<em>t - 1/2</em>9.8*t^2.</p>
<p>which gives me t = 0 and t= .41s with 2m/s for the velocity.</p>
<p>But the web homework site is telling me to try again.</p>
<p>Wouldn't you just set the equations</p>
<p>d=10t for passenger,
d=22t-4.9t^2 (22m/s relative to the ground)
and then equate and solve the quadratic equation?</p>
<p>why 4.9??? gravity is 9.8. can you explain your reasoning?</p>
<p>4.9 is from 1/2* (9.8)* t^2</p>
<p>why is the initial velocity 22???</p>
<p>We're setting the reference frame as the ground. Hence, if the balloon rises at 12m/s relative to the balloon, the speed relative to the ground will be 10+12=22m/s since the balloon is already moving at 10m/s relative to the ground and you're adding an additional velocity to that.</p>
<p>That is though, as I understand, if the the speed of 12m/s is relative to the balloon. If it was relative to the ground as stated in the question you should be able to plug in 12m/s for the initial velocity of the ball and obtain the right answer...</p>
<p>Ok, you kind of lost me. Stated as it is (I copied word for word)
The answer would be
d=10t for passenger,
d=22t-4.9t^2 (22m/s relative to the ground)</p>
<p>2.44s. Is still what you are trying to say?</p>
<p>Yes, 2.45 s is the correct answer.</p>
<p>qhausqkqh set the problem up correctly.</p>
<p>You only need two equations: one for the position of the ball, and one for the position of the balloon.</p>
<p>y(ball)= -4.9t^2 + 22t
y(balloon)=10t</p>
<p>You want to set them equal to find out when the ball comes back down and reaches the balloon again, exactly how qhausqkqh did it:
-4.9t^2 + 22t = 10t</p>
<p>You solve for t and get t=0 or t=2.45. It can't be t=0, since that's when the man threw the ball. So t=2.45 s is when the ball comes back down to the balloon.</p>