Release Height Time in Air 1.338m 0.97s A ball is tossed directly upward from th
ID: 2283183 • Letter: R
Question
Release Height Time in Air
1.338m 0.97s
A ball is tossed directly upward from the edge - timer is started as soon as the ball is released and the timer is stopped when the ball makes contact with the floor.
Find the values for : Initial vertical velocity/ final vertical velocity/ maximum height traveled by ball?
initial means just as the ball leaves the edge - final means just before the ball hits the ground. Assume no air resistance, and pay attention to your signs.
Which is greater in magnitude (ignore the sign), your initial vertical velocity or your final velocity? Is this expected why or why not?
Explanation / Answer
Ball leaves at at height 1.338 m
Spends time 0.97 s in air.
Let it gets launched up at U and touches floor at V m/s.
So, total distance travelled
- 1.338 = Ut + 0.5at2 (negative sign as ball travelling downwards)
= U*t - 0.5*9.8*t2
or - 1.338 = U*0.97 - 4.61
or 3.272 = U*0.97
or U = 3.37 m/s upward.
Maximum height:
When launched at 3.37 m/s, body travels up for distance say X.
At top point, velocity = 0 m/s
From equation of motion:
02 = 3.372 + 2*a*X
or -3.372/2+(-9.8) = X
or X = 11.356/19.6
= 0.579 m
Maximum height = height of launch + X
= 1.338 + 0.579
= 1.92 m (ans)
Final velocity
Now, V2 = U2 +2aS
V2 = U2 +2*9.8*1.338
or V2 = 3.372 + 26.2248
or V = sqrt(37.5817)
= 6.13 m/s downwards.
So, launch velocity at top = 3.37 m/s
velocity just before touching ground = - 6.13 m/s or 6.13 m/s downwards.
Maximum height reached wrt ground = 1.92 m
Launch velocity is less than velocity at ground because:
1. At launch position, ball is at a height of 1.338 metres. Hence it possesses some potential energy
equals to its mass time height * g.
2. When in motion and comes down, this potential energy is converted to kinetic energy. Additionally it has some initial kinetic energy during launch.
3. Hence, the two kinetic energies get added up resulting in higher velocity near ground.