Answer:
The height of the cliff is, h = 78.4 m
Explanation:
Given,
The horizontal velocity of the projectile, Vx = 20 m/s
The range of the projectile, s = 80 m
The projectile projected from a height is given by the formula
<em> S = Vx [Vy + √(Vy² + 2gh)] / g
</em>
Therefore,
h = S²g/2Vx²
Substituting the values
h = 80² x 9.8/ (2 x 20²)
= 78.4 m
Hence, the height of the cliff is, h = 78.4 m
Picture #1:
GPE = (mass) x (gravity) x (height)
GPE = (2 kg) x (9.8 m/s²) x (40 m) = 784 joules
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (5 m/s)²
KE = (1 kg) (25 m²/s²) = 25 joules
Picture #2:
KE = (1/2) (mass) (speed²)
KE = (1/2) (2 kg) (10 m/s)²
KE = (1 kg) (100 m²/s²) = 100 joules
Picture #3:
GPE = (mass) x (gravity) x (height)
GPE = (20 kg) x (9.8 m/s²) x (2 m) = 392 joules
KE = (1/2) (mass) (speed²)
KE = (1/2) (20 kg) (5 m/s)²
KE = (10 kg) (25 m²/s²) = 250 joules
Picture #4:
GPE = (mass) x (gravity) x (height)
98 joules = (1 kg) x (9.8 m/s²) x (height)
Height = (98 joules) / (1 kg x 9.8 m/s²)
Height = 10 meters
Picture #5:
GPE = (mass) x (gravity) x (height)
39,200 Joules = (mass) x (9.8 m/s²) x (20 m)
Mass = (39,200 joules) / (9.8 m/s² x 20 m)
Mass = 200 kg
The ball should take twice as long to return to its original position as it took to reach its maximum height, so it should return to its original position at 
.
