Question

a cart is moving horizontally along a straight line with constant speed 30m/s .a projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80m . At what speed (relative to the cart) must the projectile be fired?​

Answer 1

Answer:

$\displaystyle{u=40/3 \ m/sec}$

Explanation:

Given ;

Velocity of car is 30 / sec .

Distance covered 80 m .

Car is moving horizontally mean in the x axis .

Now total time taken by car to cover 80 m distance :

T = 80 / 30

T = 8 / 3 sec .

Now the projectile is fire from the moving .

Then both will taken same time to cover 80 m distance .

In y axis we know

T = 2 u y / a  y

T = 2 u / g

T = 2 u / 10

Putting T = 8 / 3 from ( i )

8 / 3 = 2 u / 10

u = 40 / 3 m/ sec .

Hence we get speed is 40 / 3 m / sec .

Answer 2

Answer:

HOLA MATE YOUR ANSWER-

=>To move horizontally with the cart, the projectile must be fired vertically with a flight time = x/vx = 80/30 = 2.67 s.

$\boxed{2nd \:part}$

The initial velocity v0 must satisfy

y = v0 t + at2,

with y = 0 and t = 2.67 s: thus 4.9t = v0

So, 4.9t = v0 and vo = 13.1 m/s at

θ = 900.

$\color{red}{HOPE \:IT \:HELPS\: YOU}$