Aball is thrown at a 60.0 ° angle above the horizontal across level ground. it is thrown from a height of 2.00 m above the ground with a speed of 20.0 mis and experiences no appreciable air resistance. the time the ball remains in the air before striking the ground is closest to?

3.75 seconds

Explanation:

Given:

y₀ = 2.00 m

y = 0 m

v₀ᵧ = 20.0 sin 60.0°

aᵧ = -9.8 m/s²

Find: t

y = y₀ + v₀ᵧ t + ½ aᵧ t²

0 = 2.00 + (20.0 sin 60.0°) t + ½ (-9.8) t²

0 = 2 + 10√3 t − 4.9 t²

Solve with quadratic method:

t = [ -10√3 ± √(300 − 4(-4.9)(2)) ] / -9.8

t = (-10√3 ± √378.4) / -9.8

t ≈ -0.218, 3.75

Since t > 0, the ball lands after approximately 3.75 seconds.

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