A particle is projected with a speed `10sqrt(2) ms^(-1)` and at an angle `45^(@)` with the horizontal. The rate of change of speed with respect to time at `t = 1`s is `(g = 10 ms^(-2))`

A particle is projected with a speed 10sqrt(2) making an angle 45^(@) with the horizontal . Neglect the effect of air friction. Then after 1 second of projection . Take g=10m//s^(2) .

A particle is projected with a speed 10sqrt(2) making an angle 45^(@) with the horizontal . Neglect the effect of air friction. Then after 1 second of projection . Take g=10m//s^(2) .

A particle is projected with a velocity of 10sqrt2 m//s at an angle of 45^(@) with the horizontal. Find the interval between the moments when speed is sqrt125 m//s(g=10m//s^(2))

A projectile is thrown with a velocity of 10sqrt(2) ms^(-1) at an angle of 45° with the horizontal. The time interval between the moments when the speeds are sqrt(125) ms^(-1) is (g=10 ms^(-2) )

A projectile is thrown with a velocity of 10sqrt(2) ms^(-1) at an angle of 45° with the horizontal. The time interval between the moments when the speeds are sqrt(125) ms^(-1) is (g=10 ms^(-2) )

The maximum height attained by a ball projected with speed 20 ms^(-1) at an angle 45^(@) with the horizontal is [take g = 10 ms^(-2) ]