A projectile, fired with unknown initial velocity, lands 19.4 s later on the sid

Question

A projectile, fired with unknown initial velocity, lands 19.4 s later on the side of a hill, 3080 m away horizontally and 427 m vertically above its starting point. (Ignore any effects due to air resistance.) (a) What is the vertical component of its initial velocity? m/s (b) What is the horizontal component of its initial velocity? m/s (c) What was its maximum height above its launch point? m (d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical? v = m/s ? =

Explanation / Answer

Figure out the horizontal and vertical components of the velocity on impact. The horizontal component doesn't change over time, so Vx_final = Vx_initial = 161m/sec (assuming you got part (a) correct). The vertical component changes over time by a simple formula: Vy_final = Vy_initial - gt So, plugging in your numbers: Vy_final = 116m/sec - (9.8m/sec²)(19.2sec) Once you have Vx_final and Vy_final, you can calculate V_final and ?: V_final = sqrt(Vx_final² + Vy_final²) tan(?) = Vy_final / Vx_final

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