Question

From a bridge at height 4.0 m above a river, you throw a pebble downward at an angle 25 degrees below the horizontal with initial speed 3.0m/s. (a) What is the angle of the pebble's velocity vector at the moment it hits the water? (b) How far did the pebble travel in the horizontal direction during its motion, from when you threw it until it hit the water?

Answer 1

(a)

The horizontal component of the velocity is

V_{​​​​​​X} = 3cos25

= 2.7189 m/s

The vertical component will be,

V_{​​​​​​y} = 1.2678 m/s

Now the horizontal component will remain same, only the vertical component will be changed as there is no acceleration is horizontal direction,

So the final velocity in the vertical direction will be,

V_{​​​​​​y} ²= (1.2678)²+(2*9.81*4)

V_{​​​​​​y} = 8.949 m/s

So, angle with the horizontal will be,

= 73.1 ⁰

^(B)

^{in this part we have to find the horizontal range,}

So, for that we have to find the time taken to reach the ground,

Using first equation of motion,

8.949= 1.2678+(9.81t)

t= 0.7829 seconds

So, distance travelled will be,

=(2.7189*0.7829)

2.128 m