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Sagot :
1.) Answer. 55.87m / s is the final velocity.
2.) Answer is 1.43 sec.
Mark me as brainlies please
Answer:
2.ag = -9.8 m/s assume vi = 0 m/s t = 3 s
Find:
vf = ? h= ?
Formula:
vf = vi + agt
Solution: vf = vi + agt
vf = 0 + (-9.8 m/s2 )(3 s)
vf = -29 m/s
(then if you're not considered in that answer here is the other one)
--Zero! Because in my problem, the height of the building is exactly equal to the distance the ball will drop in three seconds. At the instant in question, the ball will be partially deformed by the ground, but not yet rebounded. It will have zero velocity.
2.Use the formula of ymax = u^2/2g to solve for the value of u.
Solving for u
ymax = u^2/2g
3 m = u^2 / (2 * -9.8)
u^2 = 3 m * 19.6 m/s^2 ……. I leave you to continue the calculations. The initial velocity is more than 7 m/s.
(but also im not sure about that, here is the other answer)
--It depends, of course, on initial conditions. What’s an initial condition? For example, the height was the ball released.
In general, however, gravity problems are reversible. How fast is the ball going when it is dropped 3 m? That’s also the initial speed of the ball when tossed up.
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