Numerical: Solve by Three Equations of Motion

So the speed is 28 m/s, and since we chose positive downward and got a positive value, the direction is downward. The question asked for a velocity, not a speed, so it is important to note the direction.

v_f = 28 m/s downward

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(a) Displacement = s = 40 m
Initial velocity = u = 0
Acceleration = a = acceleration due to gravity = g = 9.8 m/s²
Time required to reach the ground = t
Equation applied : s = ut + (1/2)at²
40 = 0*t + (1/2)*(9.8)t² = 4.9t²
=> t² = 40/4.9 = 400/49
=> t = ± √(400/49) = ± 20/7 s
Rejecting the negative value, t = 20/7 s = 2.86 s
(b) Initial velocity = u = 0
Acceleration = a = acceleration due to gravity = g = 9.8 m/s²
Final velocity on striking the ground = v
Displacement = s = 40 m
Equation applied : v² = u² + 2as
v² = 0 + 2(9.8)40 = 8*98 = 16*49
v = ± √(16*49) = ± 4*7 = ± 28
Rejecting the negative value, v = 28 m/s

s=40m , u=0 . , v=? , a=g=9.8 , t=?

a) v^2 - u^2 =2as
v^2 = 2*9.8*40
v = 28m/s

also
b) v=u+at
28 = 9.8*t
t=2.8s