HW2-phy2048-Spring-2014-solution

4 23 thus the speed at impact is v v0 cos 0 2

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Unformatted text preview: /s)(3.03 s) = 758 m. (c) And from Eq. 4- 23, we find vy = gt = (9.80 m/s2 )(3.03 s) = 29.7 m/s. 4P28. (a) Using the same coordinate system assumed in Eq. 4- 22, we solve for y = h: 1 h = y0 + v0 sin θ 0t − gt 2 2 which yields h = 51.8 m for y0 = 0, v0 = 42.0 m/s, θ0 = 60.0°, and t = 5.50 s. (b) The horizontal motion is steady, so vx = v0x = v0 cos θ0, but the vertical component of velocity varies according to Eq. 4- 23. Thus, the speed at impact is v= ( v0 cosθ 0 ) 2 + ( v0 sin θ 0 − gt ) = 27.4 m/s. 2 (c) We use Eq. 4- 24 with vy = 0 and y = H: ( v0 sinθ 0 )2 = 67.5 m. H= 2g 4P32. We adopt t...
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This document was uploaded on 03/24/2014.

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