a. Find the speed required to throw a ball straight up and have it return 6 seconds later. Neglect air resistance.
b. How high does the ball go?
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see below.
it will go on earth 58.8 meters high
To solve both parts of the question, we can use the equations of motion for projectile motion. Since the ball is thrown straight up and comes back to its initial position, the total time of flight is 6 seconds.
a. To find the speed required to throw the ball, we can use the equation for the vertical velocity (v) of an object in projectile motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity and the final velocity are equal in magnitude but opposite in direction since the ball returns to its starting position. The acceleration due to gravity (g) is acting in the opposite direction of the initial velocity. So we have:
v = -u + (-g)t.
Since u and g both have negative signs, the equation becomes:
v = -(u + gt).
Since the time of flight is 6 seconds, we substitute t = 6 into the equation:
v = -(u + g(6)).
To find the speed at which the ball must be thrown, we need to find the magnitude of v. To do this, we take the absolute value of v:
|v| = |-(u + g(6))|.
|v| = u + g(6).
Therefore, the speed required to throw the ball straight up is u + g(6).
b. To find the maximum height reached by the ball, we can use the equation for displacement (s) in projectile motion:
s = ut + (1/2)at^2.
In this case, since the ball starts and ends at the same position, the displacement is zero. Therefore, we have:
0 = ut + (1/2)gt^2.
Using t = 6 (as the total time of flight) and rearranging the equation, we can solve for the initial velocity u:
-(1/2)gt^2 = ut,
-(1/2)g(6)^2 = 6u,
Simplifying the equation gives us:
-18g = 6u,
u = -3g.
The initial velocity required to throw the ball straight up is -3g.
To find the maximum height (h) reached by the ball, we can substitute the initial velocity into the equation for displacement:
s = ut + (1/2)gt^2,
s = -3g(6) + (1/2)g(6)^2,
s = -18g + 18g,
s = 0.
Therefore, the ball does not reach any height and remains at the same height as the starting position.