Respuesta :
Answer:
a) v = 19.60 m/s
b) t = 2.21 s
Explanation:
The time taken to reach the maximum height can be derived from the first equation of motion.
v = u + at ......1
given
v = 0 at the maximum height
u = 7.79 m/s
a = -g = -9.8m/s^2
Equation 1 becomes
v = u - gt =0
u = gt
t = u/g
t = 7.79/9.8 = 0.795s
t1 = 0.795s
The maximum height can be derived from;
s = ut + 0.5at^2
Since it's against gravity a = -g and s = h
The equation becomes
h = ut - 0.5gt^2
Substituting the given and derived values, we have
h = 7.79(0.795) - 0.5(9.8 × 0.795^2)
h = 3.096 m
Total maximum height from the ground = 3.096m + 16.5m = 19.596m
H = 19.596m
The final velocity of the rock when it hits the ground can be derived using the fourth equation of motion
v^2 = u^2 + 2as
Motion from the maximum height to the ground.
u = 0 (initial speed at the maximum height is 0)
a = g (acceleration due to gravity) = 9.8m/s^2
s = H = 19.596m
Substituting the values.
v^2 = 0 + 2×9.8× 19.596
v = √( 2×9.8× 19.596)
v = 19.60m/s
Using
v=u+at^2 ......3
u = 0 and a = g
From equation 3
t^2 = v/g
t =√(v/g)
v = 19.60m/s
t2 = √(19.60/9.8)
t2 = 1.414s
Total time of flight t = t1 + t2 = 0.795s + 1.414s
t = 2.209s
t = 2.21 s
