Question

A body is thrown vertically upward with speed 40 m/s. The distance traveled by the body in the last second of the upward journey is: (use g = 9.8 m/s² and ignore air resistance)

Answer 1

Using first kinematics law in vertical direction,

v - u = a*t

here, v = final speed at maximum height = 0

u = initial vertical speed = 40 m/s

a = acceleration in vertical direction = -g = -9.8 m/s^2

t = time required to reach maximum height = ??

t = (v - u)/a = (0 - 40)/(-9.8) = 4.082 s

now, distance travelled in last second will be:

d = (distance travelled till 't = 4.082 s') - (distance travelled till 't = 3.082 s')

using second kinematics law,

d = (u*t1 + 0.5*a*t1^2) - (u*t2 + 0.5*a*t2^2)

here, t1 = 4.082 s

t2 = 3.082 s

So,

d = (40*4.082 - 0.5*9.8*4.082^2) - (40*3.082 - 0.5*9.8*3.082^2)

d = 4.8964 m

In two significant figure:

d = 4.9 m

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