Question

At time t = 0, a projectile is launched from ground level. At t = 2.00 s, it is displaced d = 51 m horizontally and h 76 m vertically above the launch point, what are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant it reaches its maximum height above ground level, what is its horizontal displacement D from the launch point?

Answer 1

t = 2 sec x = 51 m as speed remains constant along horizontal so U_x = 51/2 = 25.5 m/s using y = u_y t + 1/2 g t^2 76 = u_y times 2 + 1/2 times 9.8 times (2)^2 rightarrow 2 u_y = 76 - 19.6 = 55.4 rightarrow u_y = 27.7 m/s if angle from horizontal is theta than tan theta = u_y/U_x = 27.7/25.5 = 1.086 rightarrow theta = 47.36 degree at maximum height , horizontal distance will be equal to half of Range therefor h = R/2 = (u^2 sin 2 theta/g)^1/2 h = u^2 sin 2 times 47.36 degree/2 times 9.8 = (squareroot u_x^2 + u_y^2 sin 94.72 degree/19.6 \r\n here u = squareroot u_x^2 + u_y^2 = squareroot (25.5)^2 + (27.7)^2 = squareroot 650.25 + 767.29 = squareroot 1417.54 = 37.65 m/s therefor h = (37.65)^2 sin 94.72 degree/19.6 h = 72.08 m