Find the velocity at the highest point, if it's projected with a speed 15ms^-1, in the direction 45° above the horizontal. Take g = 10 ms^-2

Open in App

Solution

At highest point the final velocity will be 15*cos45 = 10.6 m/sec in the horizontal direction (parallel to ground).
At highest point the y component of velocity becomes zero so only velocity in the horizontal direction remains.

Note that :- the velocity in the horizontal direction remains same throughout the trajectory as there is no component of force in the horizontal direction.

Suggest Corrections

Similar questions

15 m s^{- 2}

45^{\circ}

g = 10 m s^{- 2}]

15

s^{- 1}

45^{o}

g = 10 m s^{- 2}

20 m s^{- 1}

15 m

g = 10 m s^{- 2})

20 m s^{- 1}

15 m

g = 10 m s^{- 2})

[g = 10 m / s^{2}]

Join BYJU'S Learning Program