The next two questions refer to the situation below. At time t = 0, a ball is th

Question

The next two questions refer to the situation below.

At time t = 0, a ball is thrown down with initial speed vo = 25 m/s with a launch angle o = 49°, with respect to the horizontal as shown, from the edge of a vertical cliff of height ho = 51 m.

1)

What is y1, the height of the ball at time t1 defined by: vy(t = t1) = 32 vy(t = 0)

y1 = -44.3 m

y1 = 32.9 m

y1 = 64.6 m

y1 = 28.3 m

y1 = 37.4 m

2)

What is x1, the horizontal distance travelled by the ball at time t1 defined by: vy(t = t1) = 32 vy(t = 0)

x1 = 47.3 m

x1 = 31.5 m

x1 = 15.8 m

x1 = 18.1 m

x1 = 20.9 m

Explanation / Answer

v0x =v0 cos (490) = 25 cos (490) = 16.4 m/s

v0y =v0 sin (490) = 25 sin (490) = 18.9 m/s

1) vy2 -v0y2 =2as

322 - 18.92 =2 x(-9.8) x (-s)

s =34.02 m

y1 = h-s =51-34.02 = 16.98 m

2) vy =v0y +at

32= 18.9 -9.8xt

t = 1.34 s

x=v0xt = 16.4x1.34 = 21.9 m

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