1. Vo = 10m/s[70o].
Xo = 10*Cos70 = 3.42 m/s.
Yo = 10*sin70 = 9.40 m/s.
a. Y = Yo + g*Tr = 0 @ max ht.
9.40 + (-9.8)Tr = 0
Tr = 0.959 s. = Rise time.
Tf = Tr = 0.959 s. = Fall time.
T = Tr+Tf = 0.959 + 0.959 = 1.92 s. = Time in flight.
Range = Xo*T = 3.42 * 1.92 = 6.56 m.
b. Tr = 0.959 m/s, Same as part a.
h = ho + Yo*Tr + 0.5g*Tr^2.
h = 2 + 9.4*0.959 - 4.9*0.959^2 = 6.51 m. above gnd.
h = 0.5g*Tf^2 = 6.51.
4.9Tf^2 = 6.51
Tf^2 = 1.33
Tf = 1.15 s. = Fall time.
Range = Xo*(Tr+Tf) = 3.42(0.959+1.15) = 7.22 m.
The increase in range is due to the increase in fall distance and fall time(Tf).
1)A ball is thrown so that it initially travels at 10ms-1 at an angle of 70 above the horizontal
a)a simple model assumes that the ball is thrown from ground level and lands at the samelevel.Find the time of flightand range of the ball.
b)A refined model assumes that the ball is at a height of 2m when it is thrown and lands at ground level. Find the range of the ball based on this assumption.
pleasehelp with a nice explanation easy to understad so i get this
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