Question

An object is thrown upward with velocity 400 m/ sec then find the time and distance travelled by it when it's velocity becomes 100m/ sec

Answer 1

Hola mate..
..here is your answer.....

1) The object is thrown with a velocity 400 m/s , that means this velocity is the initial velocity = U

so , now , using the first Kinematical equation ,

V = U + at

but , here , a = -g

V = U - gt

V-U / g =. t

hence , time t =. 100 - 400 / -10 = 30 sec.

2) using the third Kinematical equation,

$v {}^{2} - u {}^{2} = 2as \\ \\ \: s \: = v { }^{2} -u {}^{2} \div 2a \\ \\ s = 100 {}^{2} - 400 {}^{2} \div(- 10×2) \\ \\ s = (100 + 400)(100 - 400) \div(- 20) \\ \\ s = 500 \times ( - 300) \div (-20) \\ \\ s = 7500m = displacement \\ \\ distance \: = 7500m$
hope u understood ^_^

regards ,

AdityaRocks1
#BrainlyStar