With what minimum speed must you toss a 130 g ball straight up to hit the 12 m-high roof of the gymnasium if you release the ball 1.3 m above the ground?
I already figured out the minimum speed, 14m/s I'm confused on the next part
With what speed does the ball hit the ground?
Do I use, for values of H, 10.7m or 12m or 13.3m?
It falls with zero initial velocity from the roof to the ground which is 12 meters high the way I read it.
v = sqrt (2 * 9.8 * 12)
How did you calculate the minimum speed?
Kf+Uf=Ki+Ui
.5*m*v^2 + 0=.5*m*(vi)^2 + m*9.8*1.3meter
m=mass
v= velo u want to solve for
vi= your initial velo, for you 14m/s
my values where .16kg ball, 13m ceiling, 1.7 off the ground. answer one for me was 14.88, answer 2 was 15.96m/s
To determine the speed at which the ball hits the ground, you can use the principle of conservation of energy. At the maximum height, all the initial kinetic energy is converted into potential energy. Therefore, you can equate the initial kinetic energy to the potential energy at the maximum height and then solve for the velocity at the maximum height.
Let's assume the velocity at the maximum height is v_f and the height from which the ball is released is h_0.
The potential energy at the maximum height (12m) is equal to the potential energy when the ball is released (1.3m above the ground):
mgh = mg(12m) (assuming g is the acceleration due to gravity and m is the mass of the ball)
The initial kinetic energy is given by:
KE_initial = (1/2)mv_i^2
Where v_i is the initial velocity.
Since energy is conserved, you can set the potential energy equal to the initial kinetic energy:
mgh = (1/2)mv_i^2
Canceling out the mass (m) and solving for v_i^2, we get:
v_i^2 = 2gh
Plugging in the values, with g as the acceleration due to gravity (9.8 m/s^2) and h as the total height traveled by the ball (13.3m):
v_i^2 = 2 * 9.8 m/s^2 * 13.3m
v_i = √(2 * 9.8 m/s^2 * 13.3m)
v_i ≈ 14.71 m/s
Therefore, the speed at which the ball hits the ground is approximately 14.71 m/s.