a)
consider the motion in upward direction as positive and down direction as negative
Y₀ = initial position of the stone = 20 m
v₀ = initial velocity of the stone = 10 m/s
a = acceleration = - 9.8 m/s²
Y = final position of the stone when it reach the maximum height
v = final velocity at the maximum height = 0 m/s
t = time taken to reach the maximum height
Using the equation
v² = v₀² + 2 a (Y - Y₀)
0² = 10² + 2 (- 9.8) (Y - 20)
Y = 25.1 m
also using the equation
v = v₀ + a t
inserting the values
0 = 10 + (- 9.8) t
t = 1.02 sec
b)
consider the motion in upward direction as positive and down direction as negative
Y₀ = initial position of the stone = 20 m
v₀ = initial velocity of the stone = 10 m/s
a = acceleration = - 9.8 m/s²
Y = final position of the stone when it reach the ground = 0 m
t = time taken to reach the ground
Using the equation
Y = Y₀ + v₀ t + (0.5) a t²
0 = 20 + 10 t + (0.5) (- 9.8) t²
t = 3.3 sec
. Using the value of q and E given by the problem, we find