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Summary for Time, Height and Range

10.A projecti...

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10.A projectile is projected horizontally with a velocity u.Show that its trajectory is parabolic.Obtain expression for time of flight and horizontal range.

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Q. A projectile is fixed with certain velocity

u

making an angle

θ

with the horizontal. With necessary diagram prove that the trajectory of the projectile is a parabola and also obtain an expression for

(i)Time of flight

(ii)max height reach

(iii)horizontal range of projectile

Q. A projectile is projected with some velocity at an angle

θ

from the horizontal so that its range is

R

and time of flight is

T

Q. Assertion :Using only vertical component of velocity of a projectile, time of flight can be calculated but horizontal range cannot be calculated. Reason: Time of flight depends on horizontal component and range depends on vertical component of velocity of projection.

Q. A javelin thrown into air at an angle with the horizontal has a range of

200 m

. If the time of flight is

5 s

, then the horizontal component of velocity of the projectile at the highest point of the trajectory is

Q. Two particles were projected one by one with the same initial velocity from the same point on level ground. They follow the same parabolic trajectory and are found to be in the same horizontal level, separated by a distance of 1 m, 2 s after the second particle was projected. Assume that the horizontal component of their velocities is 0.5 m s

^{- 1}

. Find
a. the horizontal range of the parabolic path.
b. the maximum height for the parabolic path.

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Summary for Time, Height and Range

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