A target lies flat on the ground 4 m from the side of a building that is 10 m tall, as shown below. The acceleration of gravity is 10 m/s^2.
Air resistance is negligible.
A student rolls a 4 kg ball off the horizontal
the roof of the building in the direction of the
target.
The horizontal speed v with which the ball
must leave the roof if it is to strike the target
is most nearly

1. v =√3 /4 m/s.
2. v =4√3 /3 m/s.
3. v = 4 m/s.
4. v = 2√2 m/s.
5. v =√2 / 4 m/s.
6. v =√5 / 4 m/s.
7. v = 4√3 m/s.
8. v = 4√5 m/s.
9. v = 6 m/s.
10. v = 4√2 m/s.

v = 2√2 m/s

Explanation:

The following data were obtained from the question:

Height (h) = 10 m

Acceleration due to gravity (g) = 10 m/s²

Horizontal distance (s) = 4 m

Horizontal velocity (v) =.?

Next, we shall determine the time taken for the ball to get to the target.

This can be obtained as shown below:

Height (h) = 10 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

10 = ½ × 10 × t²

10 = 5 × t²

Divide both side by 5

t² = 10/5

t² = 2

Take the square root of both side.

t = √2 s

Finally, we shall determine the horizontal velocity as shown below:

Horizontal distance (s) = 4 m

Time (t) = √2 s

Horizontal velocity (v) =.?

s = vt

4 = v × √2

Divide both side by √2

v = 4 /√2

Rationalise the denominator

v = (4 /√2) × √2/√2

v = 4 × √2 / √2 × √2

v = 4 × √2 / 2

v = 2√2 m/s

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

truck y has a greater velocity than truck x.