Asked by Jess
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?
Answers
Answered by
bobpursley
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
time? h=hi+vi*t-4.9t^2
solve for t
distance? 2h+50
Answered by
Lakshya
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
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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