Question

a particle is thrown upward with the speed of 39.2m/s find the time for which it moves in upward direction and maximum height it reached

Answer 1

Initial speed of the particle (u) = 39.2 m/s

Final speed at the highest point (v) = 0 m/s

Let the maximum height reached be H.

Acceleration due to gravity = - 9.8 m/s^2

Let the time taken to reach the highest point be t.

Now, using first equation of motion:

v = u + at

0 = 39.2 - 9.8 × t

$t = \frac{39.2}{9.8}$

t = 4 seconds

Now, using the third equation of motion:

$v^2 = u^2 + 2aH$

$H = \frac{v^2 - u^2}{2a}$

$H = \frac{39.2^2}{2\times 9.8}$

H = 78.4 meters