A car starts from rest and accelerates at 5 m / s 2 At t=4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t =6 s ? (Take . g =10 m / s 2)

Question

A car starts from rest and accelerates at 5 m / s 2 At t=4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t =6 s ? (Take . g =10 m / s 2) (A) 20 m / s , 5 m / s 2 (B) 20 m / s , 0 (C) 20 √2 m / s , 0 (D) 20 √2 m / s , 10 m / s 2

Tardigrade Admin · Accepted Answer

Velocity of the car at t =4 sec is V x = at =4 × 5=20 m / s So horizontal velocity =20 m / s (remain constant) Vertical velocity at t =6 sec i.e. after 2 sec of free fall Vy=g t=20 m / s So net velocity =√202+202=20 √2 m / s and once it starts falling acceleration is only ' g ' i.e., 10 m / s 2

Q. A car starts from rest and accelerates at $5 m / s^{2}$ At $t = 4 s$ , a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t = 6 s$ ? (Take $g = 10 m / s^{2})$

$20 m / s, 5 m / s^{2}$

$20 m / s, 0$

$202 m / s, 0$

$202 m / s, 10 m / s^{2}$

Q. A car starts from rest and accelerates at 5m/s2 At t=4s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t=6s ? (Take g=10m/s2)

20m/s,5m/s2

20m/s,0

202​m/s,0

202​m/s,10m/s2

Velocity of the car at t=4sec is Vx​=at=4×5=20m/s So horizontal velocity =20m/s (remain constant) Vertical velocity at t=6sec i.e. after 2sec of free fall Vy​=gt=20m/s So net velocity =202+202​=202​m/s and once it starts falling acceleration is only ' g ' i.e., 10m/s2

Q. A car starts from rest and accelerates at $5 m / s^{2}$ At $t = 4 s$ , a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t = 6 s$ ? (Take $g = 10 m / s^{2})$

$20 m / s, 5 m / s^{2}$

$20 m / s, 0$

$202 m / s, 0$

$202 m / s, 10 m / s^{2}$