From the top of a 11m. high tower a stone is projected with speed 100m/s, at an angle of 37°. Calculate speed after 2sec. and speed just before striking the ground.
Answers
Answer:
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Speed after 2 sec = 2√65 m/s = 16.12 m/s
time of flight = 2.2 sec
horizontal range = 17.6 m
the maximum height attained by the particle = 12.8 m (from ground)
speed just before striking the ground = 8√5 = 17.89 m/s
Explanation:
Fig not shown
so 37° angle not clear exactly
Assuming 37° with horizontal & upward
Velocity = 10 m/s
Horizontal Velocity = Vcos37 = 10 * 0.8 = 8 m/s
Vertical Velocity = Vsin37 = 10* 0.6 = 6 m/s
Speed after 2 sec
Horizontal Velocity remains same = 8 m/s
Vertical Velocity = 6 - 10*2 = -14 m/s => downward velocity
Velocity = √8² + 14² = 2√65 m/s = 16.12 m/s
Max height attained when vertical velocity = 0
t = 6/10 = 0.6 sec
Vertical distance covered = 6²/(2 * 10) = 1.8 m
Max height from ground = 11 + 1.8 = 12.8 m
2 * 0.6 = 1.2 sec (time of flight for 11 m height)
Vertical velocity = 6 m/s down ward
11 m vertical distance will be covered in
6t + (1/2)10t² = 11
=> 5t² + 6t - 11 =0
=> 5t² -5t + 11t - 11 = 0
=> 5t(t - 1) + 11(t - 1) = 0
=> t = 1 sec
time of flight 1.2 + 1 = 2.2 sec
Horizontal Range = 8 * 2.2 = 17.6 m
Speed just before striking the ground
Vertical Velocity = 6 - 10*2.2 = 16 m/s
Horizontal = 8 m/s
Velocity = √8² + 16² = √320 = 8√5 = 17.89 m/s