Horizontal Cannon on a Cliff (Figure 1) A cannonball is fired horizontally from

Question

Horizontal Cannon on a Cliff (Figure 1) A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 50.0m above ground level, and the ball is fired with initial horizontal speed vo. Assume acceleration due to gravity to be g = 9.80m/s2. Given that the projectile lands at a distance D = 170m from the cliff, as shown in the figure, find the initial speed of the projectile, v0. Express the initial speed numerically in meters per second. What is the y position of the cannonball when it is at distance D/2 from the hill? ff you need to, you can use the trajectory equation for this projectile, which gives y in terms of x directly: You should already know vox from the previous part. Express the position of the cannonball numerically in meters.

Explanation / Answer

Cannonball is fired horizontally from the top of cliff,Height H=50 m.

Let us consider initial velocity is zero (Vf=0).

Since the ball was fired horizontal speed assume verticalvelocity is zero (Vv=0). acceleration is 9.8 m/s2.

we plug these numbers into the equation Yf = H + Voy*t +1/2at^2

and get 0 = 50 + 0t + 1/2(-9.8)t^2

t will equal 3.194 seconds

but you want to know the height at half the time so simply dive tby 2 to get 1.597 s

now we can solve the height (we'll just call it Y) for time tg/2 byusing the same equation. replace the previously unknown t with thefound tg/2 value.

Y = 50 + 0(1.597) + 1/2(-9.8)(1.597^2)

Y = 50 - 13 = 37.5 m=    37.5 m is your height at tg/2

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