Standing on the top ledge of a 55 meter high ?

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To determine how long it takes for a ball thrown straight up from a 55-meter building to hit the ground, the formula d = h + vt + gt²/2 is used, where d is the height at time t, h is the initial height, v is the initial speed, and g is the gravity constant. For the first problem, the calculated time to hit the ground is 7 seconds. In the second scenario, where the building is 56 meters high and the ball is thrown at 30 m/s, the maximum height reached is approximately 102 meters. To find the time when the ball reaches its maximum height, the speed at that moment (s) is set to zero. This approach effectively utilizes the physics of motion under gravity to solve the problems.
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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...
 
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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.
 
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