Question

A point object is thrown vertically upwards at such a speed that it returns the thrown after 6 seconds. With what speed was it thrown up and how high did it rise? Plot speed-time graph for the object and use it to find the distance travelled by it in the last second of its journey.

Answer 1

Time of rise t₁ = time of fall = t₂

t = 6 s

v = u – g t (upward direction is chosen as positive)

v = – u

– u = u - gt

⇒ – 2u = – 9.8 × 6

⇒ u = 29.4 ms^-1

v² = u² – 2 gh

At the top, v = 0

⇒ h = u²/2g = (29.4)²/(2 x 9.8)

h = 44.1 m

Initial speed = 29.4 ms^-1

Height the object attained = 44.1 m

Distance covered in last second = area under speed – time graph.

= 1/2 × (6 – 5) × (19.6 + 29.4)

= 24.5 m