The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where should a target be placed so that the toy car lands on it?

1.6 m

Explanation:

Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.

The time for landing should be calculated by using the second equation of motion formula

h = Ut + 1/2gt^2

Let U = 0

0.5 = 1/2 × 9.8 × t^2

0.5 = 4.9t^2

t^2 = 0.5 / 4.9

t^2 = 0.102

t = 0.32 s

The target should be placed so that the toy car lands on it at:

Distance = 5 × 0.32

distance = 1.597 m

Distance = 1.6 m

Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.

this could be newton#'s law of cooling. the cooling curve is exponential