THE RANGE OF A PROJECTILE IS 1.5KMWHEN IT IS PROJECTED AT AN ANGLE OF 15 DEGREE WITH HORIZONTAl. WHAT WILL BE ITS RANGE WHEN IT IS PROJECTED AT AN ANGLE OF 45 DEGREE WITH HORIZONTAL?
Asked by shreya | 26th Oct, 2013, 08:49: PM
Let u be the velocity of projection. When the angle of projection is 150 the range is 1500 m.
So, using, R = u2 sin(2?)/g
1500 = u2 sin(2×15)/g
=> u2 = (1500 × g)/sin(30)
Now, the angle of projection is 450.
So,
R = u2 sin(2×45)/g = u2/g
= [(1500 × g)/sin(30)]/g
= 3000 m
= 3 km
Let u be the velocity of projection. When the angle of projection is 150 the range is 1500 m.
So, using, R = u2 sin(2?)/g
1500 = u2 sin(2×15)/g
=> u2 = (1500 × g)/sin(30)
Now, the angle of projection is 450.
So,
R = u2 sin(2×45)/g = u2/g
= [(1500 × g)/sin(30)]/g
= 3000 m
= 3 km
Answered by Faiza Lambe | 27th Oct, 2013, 10:02: AM
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