Question

A ball is thrown vertically upward with a velocity of 29.4m/s . Describe the motion of the ball. ​

Answer 1

Acceleration, a =g=9.8m/s

From third equation of motion, height (s) can be calculated as:

v 2 −u 2 =2gs

s=(v 2 −u 2 )/2g= ((0) 2 −(29.4) 2 )/2(−9.8)=3s

Time of ascent = Time of descent

Hence, the total time taken by the ball to return to the players hands = 3+3=6 s

hope this helps u ☺️✌❤️❤️

Answer 2

Answer:

Depends what planet youre on.

Theres a few ways to solve this problem, either use some kinematic equations, or apply conservation of energy.

Thetes 4 kinematic equations, so lets just pick the one for distance that says distance traveled is initial velocity*time + half*acceleration*time squared, or

(1) d = 29.4*t + 0.5*at^2

Note that acceleration in this case is negative, since its in the opposite direction of initial velocity.

You cant solve for d without the time, but you can find that by observing that the maximum height is reached at the time when acceleration eats up all the velocity, so use another kinematic equation to solve for time. Final velocity = initial velocity plus acceleration*time, or

(2) Vf = Vi + at

Final velocity is zero, and we know initial velocity so

0 = 29.4 + at

Keep in mind that acceleration is negative. Time cant be negative. Solve for t

t = -29.4/a

Now substitute (2)->(1)

d = (-29.4^2)/a + (29.4^2)/2a

Hmm, better combine those fractions.

d = 2*(-29.4^2)/2a + (29.4^2)/2a

Add, square, and divide, and we got

d = (-432.18)/a

Be sure to recall that acceleration is negative! Thats was a fun exercise but was kind a lot of work.

What about the other distance equation? The one that says distance is the average of initial and final velocity*time?

(3) d =[(Vi+Vf)/2]*t

So

d = [(29.4+0)/2]*t = 14.7*t

Substitute time (2)->(3)

d = 14.7*(-29.4/a) = -432.18/a

That went a little easier, keeping in mind how negative acceleration is, but if you don't wanna do the substitution, then work smart and use the time irrelevant equation; final velocity squared = initial velocity squared + twice acceleration*distance, or

(4) Vf^2 = Vi^2 +2ad

Again, again, acceleration is negative. Plug in zero as your final velocity (top of the arc) and 29.4 is your starting speed and you got

0 = 29.4^2 + 2ad

Keeping in mind acceleration is negative, Rearrange and simplify to solve for d

d = -432.18/a

Now put in your acceleration (which is negative) for your chosen planet. On earth thatll be somewhere near 43 meters, but what if youre in deep space where there's no gravity?

d=-432.18/0 - uh oh, divide by zero. I guess theres no max distance and itll fly forever.

Now for the conservation of energy method:

Kinetic energy - potential energy = 0

So 0.5*mass*velocity squared = mass*gravitational acceleration*height

0.5mv^2 = mgh

Mass cancels out, solve for h

h = (v^2)/2g

thats 432.18/g

Note, this time acceleration isnt negative. Funny that, huh? This is because we are now taking about magnitudes of energy without regards to movement direction, and you cannot have negative energy! Antienergy? An irrational concept.

So (v^2)/2g now were really working smart. Note that this equation is actually identical to all 3 of the simplified kinematic equations. Shortcutted right to the end; isnt physics elegant? Its often wise to consider all the paths to the answer before you begin working, especially if youre timed like on a test

Explanation:

$Mark Me As The Brainliest Please$