Question

13.
A Bomber flying upward at an angle of 53° with the vertical releases a bomb at an altitude of 800 m.
The bomb strikes the ground 20 s after its release. Find: [Given sin 53° -0.8; g = 10 m/s?|
(i) The velocity of the bomber at the time of release of the bomb.
(ii) The maximum height attained by the bomb.
(iii) The horizontal distance travelled by the bomb before it strikes the ground
(iv) The velocity (magnitude & direction) of the bomb just when it strikes the ground.
thc​

Answer 1

Explanation:

Given data states that the bomber flying at an angle 53∘53∘ with bomb released at an altitude of 800m800m denoted by hh and strikes ground in time t=20t=20 sec .. therefore by equation of motion,

h=ut+12at2⇒h=uyt+12gt2⇒−800=uy×20+12(−10)×(20)2⇒−800=20uy−2000⇒20uy=1200=ucos35∘h=ut+12at2⇒h=uyt+12gt2⇒−800=uy×20+12(−10)×(20)2⇒−800=20uy−2000⇒20uy=1200=ucos⁡35∘

Therefore u y is the vertical component,

uy=ucos35∘⇒60=ucos53degree⇒u=100ms