A palm fruit dropped to the ground from the top of a tree 45m tall. How long does it take to reach the ground? (g = 10ms-2)
9s
4.5s
6s
7.5s
3s
Explanation
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t= square root of 2h/g t= square root of 2*45/10 t = square root of 90/10 t = square root of 9 t = 3s
h=ut+1/2gt^2 h=45m, u=0, t=?, g=10m/s^2 45 = 0*t + 10t^2/2 45 = 10t^2/2 45*2 = 10t^2 t^2 = 90/10 t = square root of 9 t = 3s
Using the equation of the time taken for an object to reach the ground in projectile motion,
t = √(2H÷√g
Where,
H= 45m
g= 10m/s²
t= √(2×45)÷10
t=√90÷10
t= √9
t= 3seconds (3s)
your formula is not correct this question is motion under gravity and you use normal motion formula the place you use s should be h which is your height to avoid confusing we that is learning
If you're confused, check this out;
Height, h=½gt²
where h=45m
g=10m/s
and t²= the unknown.
45=½×10×t²
(2 divides 10 to give 5), so now you have
45=5t²
(5 divides 45 to give 9), and you now have
t²=9
t=√9
and what's the square root of 9?
t=3s.
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