Physic problems?

Question

) A ball has an initial speed of 23m/s. The acceleration of gravity is 9.8 m/s ^2. What will be its position after 3.5 s if it is thrown down with inital speed of 23 m/s? and up with an intitial speed of 23 m/s?

I got - 507.7450m for the first part and 668.2450 m for the second one, but I think I did them wrong. Can someone please check and correct me and tell me how to do it. thanxs

Answer 1

s = u*t + 0.5*a*t^2

Here u = 23 m/s

Thrown downwards, gravity works with the ball

s = 23*3.5 + 0.5*9.8*3.5^2 = 104.5 m below where it started from.

Thrown upwards, gravity acts against the ball. We need to find out how long it takes to reach the maximum height, after which it starts to fall back down. (at max height, v = 0)

v = u + a*t

0 = 23 - 9.8*t

t = 2.3 s.

Therefore it is on it's way back down at t = 3.5 s

Let's calculate how high it climbs

v^2 = u^2 + 2as

0 = 23^2 - 2*9.8*s

s = 27 m.

Now let's think about the trip down from the highest point. It took 2.3 s to get there, meaning that it is coming back down for 3.5 - 2.3 = 1.2 s. It starts with u = 0 and gravity is helping it.

s = ut + 0.5*a*t^2

s = 0 + 0.5*4.9*1.2^2

s = 7.1 m (this is how much it has come back down)

So the total height above the starting point is 27 - 7.1 = 19.9 m.

Answer 2

use the equation S= ut + 1/2 at^2
for the first one you take th acceleration to be positive as the motion and the direction of acceleration is same
and for second you take it to be negetive as dirction of motion is opp the direction of acceleration
so for first the ans will be 140.525
and for second the ans is 20 .475
hope that helps

Answer 3

h=-1/2gt^2+Vot
!)h=-4.9*(3.5)^2-23*3.5=-140.525m (the - sing means below its initial position)
2)h=-4.9*(3.5)^2+23*3.5=20.475m (above initial position)
The units are meter