Answer:
a) 256 ft
b) 32 ft/s
c) -32 ft/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32 ft/s²
Maximum height = 256 ft
When s = 240 ft
Speed of the ball at 240 ft is 32 ft/s while going up
Now this will be the velocity of the ball when it reaches 240 ft, which will be considered as the initial velocity
Now, initial velocity will be considered as zero
Speed of the ball at 240 ft is -32 ft/s while going down
Answer:
v₁
Explanation:
A crew member is walking on a tugboat that is pulling a barge. The tugboat is moving at a constant speed upstream in a river that has a constant downstream current. The velocity of the crew member with respect to the tugboat is v₁. What is the velocity of the crew member with respect to the barge?
velocity is the change in displacement per time.
the velocity of the crew member with respect to the barge is v₁ since the barge is moving in the velocity of tugboat
Explanation:
For projectile motion, use constant acceleration equation:
Δx = v₀ t + ½ at²
where Δx is the displacement,
v₀ is the initial velocity,
a is the acceleration,
and t is time.
Both objects are projected upward with velocity u. The second object is thrown after a time t₀.
For the first object:
Δx = u t + ½ (-g) t²
Δx = u t − ½g t²
For the second object:
Δx = u (t−t₀) + ½ (-g) (t−t₀)²
Δx = u (t−t₀) − ½g (t−t₀)²
Assuming the objects meet, the displacements will be equal:
u t − ½g t² = u (t−t₀) − ½g (t−t₀)²
u t − ½g t² = u (t−t₀) − ½g (t² − 2tt₀ + t₀²)
u t − ½g t² = u t − u t₀ − ½g t² + g tt₀ − ½g t₀²
0 = -u t₀ + g tt₀ − ½g t₀²
0 = -u + g t − ½g t₀
g t = u + ½g t₀
t = u/g + t₀/2
the answer for this question is c
Answer:
Final velocity = 16 m/s
Total distane travelled = 390 m
Explanation:
We can use equation of motion to solve this:
