A projectile is fired with an initial speed of 34.0 m/s at an angle of 35.0^\\ci

Question

A projectile is fired with an initial speed of 34.0 m/s at an angle of 35.0^circ above the horizontal. The object hits the ground 9.50 s later.
a)How much higher or lower is the launch point relative to the point where the projectile hits the ground?
B)To what maximum height above the launch point does the projectile rise?
C)What is the magnitude of the projectile's velocity at the instant it hits the ground?
D)What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?

Explanation / Answer

let the initial vertical position be H. a) we have y=H+34*sin35*t-g*t^2/2 when t=9.5, y=0. so 0=H+34*sin35*9.5-g*9.5^2/2 so H=257(m) so its 257m higher. b) y max when y'=0. so y'=34*sin35-g*t=0. so t=2(s). so y_max=276.4(m). so distance from initial position. 276.4-257=19.4(m) c) conservation of energy. 34^2/2+257*g=v^2/2 so v=78.7(m/s) d) horizontal component is the same = 34*cos35=27.9 so cos alpha = 27.9/78.7 so alpha = 69(deg)

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