Answer :
Answer:
Velocity before hitting the bottom = 280m/s
Explanation:
m = mass of car = 1380kg, u = initial velocity = 0,
H = height of the hill = 28m, v = final velocity = ?
From equation of vertical motion,
[tex] {v}^{2} = {u}^{2} + 2gh[/tex]
[tex] {v}^{2} = 0 + 2 \times 10 \times 28 = 280 \: m {s}^{ - 1} [/tex]
Velocity before hitting the bottom = 280m/s
Answer:
23.4 m/s
Explanation:
Energy is conserved. The car starts at the top of the hill with only gravitational potential energy (PE). When it reaches the bottom of the hill, all of the energy has been converted to kinetic energy (KE).
PE = KE
mgh = ½ mv²
v² = 2gh
v² = 2 (9.8 m/s²) (28.0 m)
v = 23.4 m/s
