A cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 100 m above ground level, and the ball i
s fired with initial horizontal speed v0. Assume acceleration due to gravity to be g = 9.80 m/s2 .a. Assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. What is the y position of the cannonball at the time tg/2?b. Given that the projectile lands at a distance D = 200m from the cliff. Find the initial speed of the projectile, v0.c. What is the y position of the cannonball when it is at distance D/2 from the hill? If you need to, you can use the trajectory equation for this projectile, which gives y in terms of x directly:
1 answer:
Answer:
a) At tg/2 the cannonball is 25 m below the launching point (75 m above the ground).
b) The initial speed of the projectile is 44.2 m/s.
c) When the x position of the cannonball is 100 m, the y position is 25 m below the launching point (75 m above the ground).
Explanation:
Hi there!
The equation of the position vector of the cannonball is the following:
r = (x0 + v0 · t, y0 + v0y · t + 1/2 · g · t²)
Where:
r = position vector of the ball.
x0 = initial positon of the ball.
v0 = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity.
Let´s place the origin of the frame of reference at the launching point so that x0 and y0 = 0.
When the ball reaches the ground at t = tg, the vertical component of the position vector will be -100 m (see attached figure). Since the cannonball is fired horizontally, it has no initial vertical velocity, then, v0y = 0. Then, the equation of the vertical component of the position vector at t = tg will be:
-100 m = 1/2 · g · tg²
Solving for tg:
2 · (-100 m / g) = tg²
√(-200 m / -9.80 m/s²) = tg
tg = 4.52 s
At t = tg/2
y = 1/2 · g · tg²/4
y = 1/2 · g · (-200 m / 4 · g)
y = -25.0 m
At tg/2 the cannonball is 25 m below the launching point (75 m above the ground).
b) Using the equation of the horizontal component of the position vector, we can calculate the initial velocity:
x = x0 + v0 · t (x0 = 0)
at t = tg, x = 200 m
200 m = v0 · 4.52 s
200 m / 4.52 s = v0
v0 = 44.2 m/s
The initial speed of the projectile is 44.2 m/s
c) Let´s calculate the time at which the ball is at an horizontal position of x = 100 m.
x = v0 · t
x/v0 = t
100 m / 44.2 m/s = t
t = 2.26 s (notice that this time is tg/2)
We have already calculated the y position at that time:
y = -25.0 m
When the x position of the cannonball is 100 m, the y position is 25 m below the launching point (75 m above the ground).
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