Physics projectile formula
A 0.11 kg ball is thrown straight up into the air with an initial velocity of 26 m/s
Find the momentum of the ball halfway to its maximum height on the way up.
V0 = 26 m/s
g=9.8 m/s^2
Max Height = V^2/2g
= 26^2 / 2*9.8
= 676 /19.6
= 34.489 meters
Halfway up is 1/2 this number or 17.2445 meters
Use the equation for height in terms of time
and solve for t
Height = Vt – 1/2 gt^2
17.2445 = 26t -1/2 (9.8)t^2
4.9t^2-26t+17.2445=0
Use the quadratic formula
( -b +- SQRT(b^2-4ac))/2a
to solve for roots
Roots are 0.777(ball on way up) and 4.52(ball on way down)
Calculate the velocity at the point when the ball is halfway to top
Use equation for velocity
V = V0 -gt
V= 26 – 9.8(0.7777)
V= 18.3854 m/s
Momentum is the product of velocity and mass
Mball x Velocity = 0.11Kg x 18.3854 m/s = 2.022 Kg *m/s
The ball will finally hit the ground 5.306 seconds after the toss
The maximum height is reached at 2.6485 seconds where the velocity is zero
Because the ball is moving fastest when first tossed, it reaches the halfway point in only 0.77 seconds but takes until 2.6 seconds to reach the maximum height.