a ship is 690 m away from port. A shell. is fired from the port with initial velocity of v0 at an elevation of 45° with the horizontal. the shell hits the ship. Initial velocity of shell is (assume ship is stationary and g=10m/s2)
Answers
Answer:
Explanation:
first ,we can find the time of flight
and we know the equation of time of flight is T=2v₀sinФ/g=2×v₀sin45/10
⇒(2×v₀×1/√2)/10⇒ T=√2v₀/10(equation 1)
we know that 690 m is the horizontal distance from the port and the horizontal component of velocity is v₀cos45=v₀×1/√2=v₀/√2
we know horizontal range(R)=v₀cosФ×T⇒ 690=(v₀/√2)×T(equation 2)
substituting equation 1 in equation 2
690=(v₀/√2)×√2v₀/10⇒ 690=v₀²/10
⇒v₀²=6900⇒ v₀=83.07m/s
hope this helps you
thanks
Initial velocity of shell is 10√69 m/s or about 83.066m/s
Given:
- A ship is 690 m away from port.
- Initial velocity of v₀ at an elevation of 45° with the horizontal
- Hit the ship which is stationary
- g = 10 m/s²
To Find:
- Initial velocity of shell v₀
Solution:
Step 1:
Find Vertical and horizontal component of velocity
vertical v₀sin45° = v₀/√2
Horizontal v₀cos45° = v₀/√2
Step 2:
Use V = U + at and find t to reach at top where v= 0
0 = v₀/√2 + gt
=> 0 = v₀/√2 + (-10)t
=> t = v₀/10√2
Step 3:
Total Time of flight is double to reach top
= 2t
= 2 v₀/10√2
= √2v₀/10
Step 4:
Find Distance using Horizontal speed x Time and equate with 690
(v₀/√2 ) (√2v₀/10) = 690
=> v₀² = 6900
=> v₀ = 10√69 m/s
v₀ ≈ 83.066m/s
Initial velocity of shell is 10√69 m/s or about 83.066m/s