A projectile is launched at ground level with an initial speed of 52.0 m/s at an

Question

A projectile is launched at ground level with an initial speed of 52.0 m/s at an angle of 31.0 degree above the horizontal. The projectile lands on a hillside 3.55 later. Neglect air friction.(Assume that the 4 x - axis is to the right and the + y - axis is up along the page.) What is the projectile's velocity at the highest point of its trajectory? Magnitude ______ m/s direction ______ degree counterclockwise from the + x - axis What is the straight-line distance from where the projectile was launched to where it hits its target? _______ m

Explanation / Answer

time at which the projectile reaches its highest point can be found using

vy = uy - gt = 0

52sin31 - 9.8t = 0

t = 2.733 s

and velocity of the projectile will be: v = 52cos31 = 44.573 m/s

b] at t = 3.55 s

x = (ucos31)t = 52cos31 (3.55) = 158.233 m

and y = uyt - (1/2)gt2 = (52sin31)(3.55) - (4.9)(3.55)2 = 33.324 m

so, the shortest distance from the launching point =

d = [x2 + y2]1/2 = [(158.233)2 + (33.324)2]1/2 = 161.704 m.

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