Question

1. A cannon is placed at the edge of a cliff 44.1m tall and tilted down by 30°. The ball is fired with an initial speed of 78.4m/s and ends up striking the ground in the level valley below the cliff. Find the range.

2. A sealed tank has 2 vertical meters of water in it and a gauge pressure of 30.4 kPa above it. There is a tiny leak at the bottom. Find the speed of the water exiting the leak.

3. A cannon is tilted up at 30° and fired at 98 m/s over level ground. The ball strikes the side of a building at a height of 78.4 m while still rising. What is the range?

Answer 1

1.

Range of projectile motion is given by:

R = V0x*T

V0x = Initial horizontal velocity = V0*cos A = 78*cos 30 deg = 67.55 m/sec

T = time period of motion = ?

to find the time period, Using 2nd kinematic equation in vertical direction

h = V0y*T + (1/2)*ay*T^2

ay = acceleration due to gravity = -9.81 m/sec^2

V0y = Initial vertical velocity = -78.4*sin 30 deg = -39.2 m/sec (-ve sign because cannon is tilted down)

h = Vertical distance traveled = -44.1 m

So

-44.1 = -39.2*T - 0.5*9.81*T^2

4.905*T^2 + 39.2*T - 44.1 = 0

Solving above quadratic equation:

T = [-39.2 + sqrt (39.2^2 + 4*4.905*44.1)]/(2*4.905) = 0.99989 = 1.00 sec

See that we have only used +ve root of equation, Since time cannot be negative.

So, Now range of baseball will be

R = V0x*T

R = 67.55*0.99989 = 67.54 m

R = range = 67.5 m (in three significant numbers)

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