Question

A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on top of the cliff 7.6 s after it is fired, find the initial velocity of the projectile. Neglect air resistance.

Answer 1

Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

$x=ucos\alpha t-(1/2)gt^{2}$ but the acceleration in the horizontal path is zero, hence we have

$x=ucos\alpha t$

Since the horizontal distance covered is 155m at 7.6secs, we have $ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1$

Also from the vertical path, the distance covered is expressed as

$y=usin\alpha t-(1/2)gt^{2}$

since the horizontal distance covered in 7.6secs is 195m, then we have

$y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2$

Hence if we divide both equation 1 and 2 we arrive at

$\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}$

Hence if we substitute the angle into the equation 1 we have

$ucos72.02=20.38\\u=66.02m/s$

Hence the initial velocity is 66.02m/s