Standing on the top ledge of a 55 meter high ?

[xterminal01]

Standing on the top ledge of a 55 meter high building you throw a ball straight up with an initial speed of 29 m/s. How long, to the nearest second, does it take to hit the ground?

If the building in the previous question was 56 meters high and you throw the ball at 30 m/s, how high to the nearest meter does it go?

The correct answer to problem number one is 7
And number two is 102

Can't figure the formulas that need to be used in order to solve these..
Any help would be appreciated...

 

[mathman]

General formula:

d=h + vt + gt²/2.
where d is height at time t.

h=initial height
v=initial speed (+ is up)
g=gravity constant (use - since it is downward) = (approx) 9.8 meters/sec²

When object hits the ground d=0, you need to solve for t to get time.

s(the speed at time t) = v+gt. To find max height, find t when s=0 and compute d for this t.

 

[kerimek]

I can provide a mathematical explanation for solving these problems. The first problem can be solved using the formula h = h0 + v0t - 1/2gt^2, where h is the height, h0 is the initial height, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the given values, we get 0 = 55 + 29t - 4.9t^2. Solving for t, we get t = 7 seconds.

For the second problem, we can use the formula v^2 = v0^2 - 2gh, where v is the final velocity and h is the height. We know that the final velocity is 0 when the ball hits the ground, so we can solve for h. Plugging in the given values, we get 0 = 900 - 2(9.8)(h), and solving for h, we get h = 102 meters. This is the maximum height the ball reaches before falling back to the ground.