Answer:
89·432 m
Step-by-step explanation:
Initial velocity of the coconut = 28 m/s
Initial velocity of the coconut will be the initial horizontal velocity of the coconut
As there is no horizontal acceleration, the horizontal velocity of the coconut remains constant
Therefore at any instant the horizontal velocity of the coconut = 28 m/s
Initial height of the coconut above the ground = 50 m
Let the time taken by the coconut to reach the ground be t s
From the formula
s = u × t + 0·5 × a × t²
where
s is the displacement
u is the initial velocity
t is the time taken
a is the acceleration
In this case taking upward direction as positive and considering the equation in vertical direction
s = -50
u = 0 (∵ initial velocity in vertical direction is 0)
a = - g = - 9·8 m/s²
- 50 = - 0·5 × 9·8 × t²
t² = 100 ÷ 9·8
∴ t ≈ 3·194 s
In this time distance travelled in horizontal direction = 28 × 3·194 = 89·432 m
∴ Horizontal distance travelled by the coconut from the point where it is dropped = 89·432 m