Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Answer:
1.) U = 39.2 m/s
2.) t = 4s
Explanation: Given that the
height H = 78.4m
The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2
Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0
Velocity of projections can be achieved by using the formula
V^2 = U^2 - 2gH
g will be negative as the object is moving against the gravity
0 = U^2 - 2 × 9.8 × 78.4
U^2 = 1536.64
U = sqrt( 1536.64 )
U = 39.2 m/s
The time it takes to reach its highest point can be calculated by using the formula;
V = U - gt
Where V = 0
Substitute U and t into the formula
0 = 39.2 - 9.8 × t
9.8t = 39.2
t = 39.2/9.8
t = 4 seconds.
Via half-life equation we have:

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

There is 6.25 grams left of Ra-229 after 12 minutes.