Question

An object is projected vertically upwards at time t = 0. Air resistance is negligible. The object passes the same point above its starting position at times 2 s and 8 s. If g = 10 m s , what is the initial speed of the object?

Answer 1

Answer: 50ms^-2

Explanation:

v=u+at

because objects travel in a parabola you find the time it reaches the vertex

which is 5 sec (10/2)

at the vertex the velocity is 0ms^-1

0=u+ (-10)(5)

0=-u-50

u=50ms^-1

Answer 2

Given: just above starting point is a point reached after 2 s and again crossed after 8s from the beginning of the journey.

g = 10m/s

To Find: initial speed = u =?

Solution:

• As the point is crossed 2 times ∴ the body went up and came down covering the same distance "d" 2 times.

• A maximum height is reached in the middle of the journey where the velocity of the object = 0

• The distance d will be covered after 2 s point will take 8/2 s.
• Similarly, the 2nd half of the journey to reach the 2s point will take 8/2 s = 4s.
• Hence total time to reach the highest point where velocity = 0 as the object is traveling against the gravitational force = 4 + 2 = 5s.

Calculating the initial velocity of the journey using Newton's first law of motion:

Time taken - t = 5s, u = ? , v - final velocity= 0, g = -10 (going upwards against the direction of gravitational force hence negative acceleration)

v = u + gt

0 = u - 10 × 5

u - 50 = 0

u = 50m/s²