# Physics

A stone is thrown vertically upwards from the top of a tower 50.0m high with an intial velocity of 20.0m/s. (a) whats the maximum height the stone reaches? (b) Whats the time it takes to reach the maximim height? (c) What is the total distance it covers?

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1. vf^2=2gh where h is the height from the tower. for total height, add 50m

time? h=hi+vi*t-4.9t^2
solve for t

distance? 2h+50

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bobpursley
2. Initial velocity=20 m/s
Final velocity=0m/s
Height using third equation of motion 2as=v²-u²
Acceleration due to gravity = -9.8 m/s(-10m/s)
- sign for acceleration in opposite direction
2*-10*s=0²-20²
-20s=-400
S=-400/-20=20metres
So maximum height= height of tower + height of stone from tower
50+20=70 m
b) time taken using first equation of motion v=u+at
0=20+(-10t)
-20=-10t
t=-20/-10 =2 seconds

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