1)A hanglider flies with a horizontal velocity of 13 m/s when his wallet falls out. If the hanglider is 1450 m above the earth when the wallet falls out, find the horizontal distance the wallet travels before hitting the ground.

The wallet is falling from a height of 1450 m

We are given that:

downward acceleration (a) = 10 m/s/s

downward displacement (s) = 1450 m

downward initial velocity (u) = 0 m/s (Since the wallet was dropped, it had no initial velocity)

horizontal initial velocity (v) = 13 m/s

Since there is no external force being applied on the wallet except the force of gravity, the wallet will keep moving at a constant velocity of 13 m/s

To calculate the horizontal distance travelled by the wallet, we need to know how long the wallet was airborne

To calculate how long the wallet was airborne, we have to find out that how long it took the wallet to hit the ground

Time taken for the wallet to hit the ground:

From the second equation of motion:

s = ut + 1/2 at² ------------(for vertical motion of the wallet)

1450 = (0)(t) + 1/2 (10)(t)²

1450 = 5t²

t² = 290

t = 17 (approx)

Time taken for the wallet to hit the ground = 17 seconds

Horizontal distance travelled by the wallet:

Since the wallet hits the ground after 17 seconds, it will move horizontally at a constant velocity of 13 m/s for 17 seconds

horizontal acceleration = 0 m/s/s

From the second equation of motion:

s = ut + 1/2at²

s = ut + 1/2 (0)t²

s = ut

here, u is the horizontal initial velocity of the wallet and the time taken by the wallet to hit the ground is 't'

s = (13)(17)

s = 221 m

Hence, the horizontal distance travelled by the wallet while falling from a height of 1450 m is 221 m

it is a good conductor

explanation: