A ball is thrown upward from the ground with an initial speed of 24.2 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground

0.66secs

Explanation:

Given the following

Initial speed u = 24.2m/s

Height H = 18m

Required

Time t

Using the equation of motion:

H = ut+1/2gt²

18 = 24.2t+1/2(9.8)t²

18 = 24.2t+0.5(9.8)t²

18 = 24.2t+4.9t²

Rearrange

4.9t²+24.2t -18 = 0

Using the general formula

t = -24.2±√24.2²-4(4.9)(-18)/2(4.9)

t = -24.2±√585.64+352.8/9.8

t = -24.2±√938.44/9.8

t=-24.2+30.63/9.8

t= 6.43/9.8

t = 0.66secs

The ball will be at the same height after 0.66secs

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

anchor = archer //