A soccer player kicks a rock horizontally off a 39.4 m high cliff into a pool of

Question

A soccer player kicks a rock horizontally off a 39.4 m high cliff into a pool of water. If the player hears the sound of the splash 3.04 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

What I have done: v((2 * delta y)/a) so v((2*39.4)/(9.8)) ˜ t˜2.84s 3.04 - 2.84 = 0.20s 343 * 0.20 gives me a resultant vector of 68.6m resultant2 - delta y2 = delta x2
so 68.62 - 39.42 =delta x2
so delta x2 = 3153.6
therefore delta x = 56.156 ˜ 56.2m delta x = initial velocity * t + (1/2)(a)(t2 )
so 56.2 = initial velocity (2.84) + (1/2)(9.8)(2.842)
when I work this out I get an initial velocity of 5.88m/s, and the homework website says it is incorrect...where am I going wrong?
What I have done: v((2 * delta y)/a) so v((2*39.4)/(9.8)) ˜ t˜2.84s 3.04 - 2.84 = 0.20s 343 * 0.20 gives me a resultant vector of 68.6m resultant2 - delta y2 = delta x2
so 68.62 - 39.42 =delta x2
so delta x2 = 3153.6
therefore delta x = 56.156 ˜ 56.2m delta x = initial velocity * t + (1/2)(a)(t2 )
so 56.2 = initial velocity (2.84) + (1/2)(9.8)(2.842)
when I work this out I get an initial velocity of 5.88m/s, and the homework website says it is incorrect...where am I going wrong?

Explanation / Answer

what you did your procedure is correct bu last part to   find   intial speed    just   you use             v = x/t1 = 56.2m/2.84s = 19.78m/s

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