Answer:
66.02m/s
Explanation:
the equation describing the distance covered in the horizontal direction is
[tex]x=ucos\alpha t-(1/2)gt^{2}[/tex] but the acceleration in the horizontal path is zero, hence we have
[tex]x=ucos\alpha t[/tex]
Since the horizontal distance covered is 155m at 7.6secs, we have [tex]ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1[/tex]
Also from the vertical path, the distance covered is expressed as
[tex]y=usin\alpha t-(1/2)gt^{2}[/tex]
since the horizontal distance covered in 7.6secs is 195m, then we have
[tex]y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2[/tex]
Hence if we divide both equation 1 and 2 we arrive at
[tex]\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\[/tex]
Hence if we substitute the angle into the equation 1 we have
[tex]ucos72.02=20.38\\u=66.02m/s[/tex]
Hence the initial velocity is 66.02m/s
