Question

(Figure
1) A cannonball is fired
horizontally from the top of a cliff. The cannon is at
height H = 80.0m
above ground level, and the ball is fired with
initial horizontal speed v0.
Assume acceleration due to gravity to be g = 9.80m/s2
.

Part
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?

Answer numerically in units of
meters.

Answer #1

**
The time it takes a cannonball fired horizontally from a 100m high
cliff to reach the ground is just**

**
H =
Vyo
* t + 1/2 * g * tg^2; (Vyo = 0 for a horizontal shot, solve for
tg)**

**
tg^2 = 2 * H / g = 200 / 9.8**

**
tg = 4.5 sec**

**
After half of this time (tg/2 = 2.26 sec), the cannon ball has
dropped:**

**
Y = Vyo * t + 1/2 * g * (2.26)^2**

**
Y = 1/2 * 9.8 * (2.26)^2**

**
Y = 25 m (from the top of the cliff)**

**
Notice that the given horizontal initial velocity Vo is irrelevant
to the problem.**

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