Answer:
a) 0.60 m
b) 2.5 m
Explanation:
Use conservation of energy to find the speed at the end of the track.
KE = PE + KE
½ mv² = mgh + ½ mv²
½ v² = gh + ½ v²
½ (5.4 m/s)² = (9.8 m/s²) (0.40 m) + ½ v²
v = 4.62 m/s
a. Use projectile motion to find the maximum height.
Given:
v₀ = 4.62 m/s sin 48° = 3.43 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (3.43 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 0.60 m
b. Use projectile motion to find the time it takes to land.
Given:
Δy = -0.40 m
v₀ = 4.62 m/s sin 48° = 3.43 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-0.40 m = (3.43 m/s) t + ½ (-9.8 m/s²) t²
4.9t² − 3.43t − 0.40 = 0
t = [ 3.43 ± √((-3.43)² − 4(4.9)(-0.40)) ] / 2(4.9)
t = (3.43 ± 4.43) / 9.8
Since t > 0:
t = 0.802
Now find the horizontal distance traveled in that time.
Given:
v₀ = 4.62 m/s cos 48° = 3.09 m/s
a = 0 m/s²
t = 0.802 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (3.09 m/s) (0.802 s) + ½ (0 m/s²) (0.802 s)²
Δx = 2.5 m