1. A stone is thrown vertically upward with an initial velocity of 40m/s.Draw the velocity time graph of the motion of stone till it comes back on the ground.
i )Find the total distance covered and net displacement by the stone.
ii )Use the graph to find the maximum height reached by the stone
Answers
Given
A stone is thrown vertically upward
From A to B
Initial Velocity (u) = 40 m/s
Final Velocity (v) = 0
Time (t) = ?
Acceleration (g) = -10 m/s²
Using first equation of motion
v = u - gt
0 = 40 - 10×t
10t = 40
t = 4 sec
Using second equation of motion
Maximum height, h = ut - 1/2×g×t²
= 40×4 - 1/2×10×4×4
= 160 - 80
= 80 m
Distance Covered = A to B + B to A = 80 + 80 = 160 m
Net Displacement = 0 (As it comes to its initial point)
Points For v-t Graph,
When, t = 0 , v = 40 ⇒ (0,40)
When , t = 4 , v = 0 ⇒ (4,0)
Similary after 4+4 = 8 second stone will touch the ground and its velocity will be 40m/sec
Now,
Using graph,
Maximum height of the stone will be given by area under the v-t graph covered by ABC
Area of ΔABC = 1/2 × 40 × 4 = 80
Maximum height = 80 m
