A projectile is shot from a cannon at an angle of 42° above the horizontal. When the cannon ball is launched, it leaves the barrel moving at 160 m/s at a height of 3.5 m above the ground.
As the cannonball leaves the barrel, what is its horizontal velocity (vx)? (Keep three significant figures.)
As the cannonball leaves the barrel, what is its vertical velocity (vy)? (Keep three significant figures.)
After the cannon ball leaves the barrel, what is its horizontal acceleration (ax)?
After the cannon ball leaves the barrel, what is its vertical acceleration (ay)?
What is the maximum altitude (ymax) attained by the cannonball? (Keep two significant figures.)
What is the range (x at impact) of the cannonball? (Keep three significant figures for intermediate calculations and round the final answer to two significant figures.)
Physics Help?
2020-06-24 4:54 am
回答 (2)
2020-06-24 9:06 pm
Considering vertical motion we have:
a(t) = -g
v(t) = -gt + v(0)
so, v(t) = -9.8t + 160sin42°
Then, s(t) = -4.9t² + 160tsin42° + s(0)
so, s(t) = -4.9t² + 160tsin42° + 3.5
With horizontal motion we have:
distance = 160tcos42°
Horizontal velocity = 160cos42° = 119 ms⁻¹
Vertical velocity = 160sin42° = 107 ms⁻¹
Horizontal acceleration is zero.
Vertical acceleration is -9.8 ms⁻²
Maximum altitude occurs when v(t) = 0
i.e. -9.8t + 160sin42° = 0
so, t = (160sin42°)/9.8 = 10.9 seconds
Hence, s(10.9) = -4.9(10.9)² + 160(10.9)sin42° + 3.5
i.e. max altitude = 590 metres
The cannonball reaches the ground when s(t) = 0
i.e. -4.9t² + 160tsin42° + 3.5 = 0
or, 4.9t² - 160tsin42° - 3.5 = 0
From the quadratic formula we get:
t = [160sin42 ± 107.38]/9.8 = 21.9 seconds
so, distance (range) = 160(21.9)cos42° = 2600 metres
:)>
a(t) = -g
v(t) = -gt + v(0)
so, v(t) = -9.8t + 160sin42°
Then, s(t) = -4.9t² + 160tsin42° + s(0)
so, s(t) = -4.9t² + 160tsin42° + 3.5
With horizontal motion we have:
distance = 160tcos42°
Horizontal velocity = 160cos42° = 119 ms⁻¹
Vertical velocity = 160sin42° = 107 ms⁻¹
Horizontal acceleration is zero.
Vertical acceleration is -9.8 ms⁻²
Maximum altitude occurs when v(t) = 0
i.e. -9.8t + 160sin42° = 0
so, t = (160sin42°)/9.8 = 10.9 seconds
Hence, s(10.9) = -4.9(10.9)² + 160(10.9)sin42° + 3.5
i.e. max altitude = 590 metres
The cannonball reaches the ground when s(t) = 0
i.e. -4.9t² + 160tsin42° + 3.5 = 0
or, 4.9t² - 160tsin42° - 3.5 = 0
From the quadratic formula we get:
t = [160sin42 ± 107.38]/9.8 = 21.9 seconds
so, distance (range) = 160(21.9)cos42° = 2600 metres
:)>
2020-06-24 6:26 am
1/
vx = 160cos42
2/
vy=160sin42
3/
ax=0
4/
ay=-g=-9.8
you can do 5 / 6/
vx = 160cos42
2/
vy=160sin42
3/
ax=0
4/
ay=-g=-9.8
you can do 5 / 6/
收錄日期: 2021-05-01 22:04:04
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