A projectile is fired Horizontally from a tower with initial velocity 50m/s .Find the height of the tower ,time of flight,Range ,if the projectile hit the ground at 45° angle with the horizontal.
Answers
Answer:
Along the y-direction,
Initial velocity = u sinθ
Acceleration due to gravity = -9.8 m/s⁻²
∴
⇒ 50 = -usinθt + 1/2 × 10 × t²
⇒ -15t + 5t² = 50
⇒ t² - 3t - 10 = 0
∴ t² - 5t + 2t - 10 = 0
⇒ t(t - 5) + 2(t - 5) = 0
∴ (t + 2)(t - 5) = 0
⇒ t = -2 and t = 5
Hence, Time is 5 seconds.
Therefore, distance along horizontal = ucosθ × t = 30 × √3/2 × 5 = 75√3 m.
For Maximum height,
H = u²sin²θ/2g
= 900/80
= 45/4 m.
= 11.25 m.
Total maximum height = 50 + 11.25 = 61.25 m.
- Read more on Brainly.in - https://brainly.in/question/4893465#readmore
Answer:
Calculating the time taken by the projectile to reach the ground:
s=ut+
2
1
gt
2
500m=(100m/s)t+
2
1
(9.8m/s
2
)t
2
500=100t+4.9t
2
t=4.15s
Calculating the distance of the target from the hil:
Range = ut=(100m/s)(4.15s)=415m
Calculating the velocity with which the projectile hits the ground:
v=100m/s+(9.8m/s
2
)(4.15s)
2
=268.78m/s