Velocity of a projectile at height `15 m` from ground is `v=(20hati+10hatj)m//s`. Here `hati` is in horizontal direction and `hatj` is vertically upwards. Then
Speed with which particle is projected from ground is……… m/s

Velocity of a projectile at height 15 m from ground is v=(20hati+10hatj)m//s . Here hati is in horizontal direction and hatj is vertically upwards. Then Angle of projectile with ground is

Velocity of a projectile at height 15 m from ground is v=(20hati+10hatj)m//s . Here hati is in horizontal direction and hatj is vertically upwards. Then Angle of projectile with ground is

Velocity of a projectile at height 15 m from ground is v=(20hati+10hatj)m//s . Here hati is in horizontal direction and hatj is vertically upwards. Then Maximum height from a ground is …….. m

Velocity of a projectile at height 15 m from ground is v=(20hati+10hatj)m//s . Here hati is in horizontal direction and hatj is vertically upwards. Then Maximum height from a ground is …….. m

Velocity of a projectile at height 15 m from ground is v=(20hati+10hatj)m//s . Here hati is in horizontal direction and hatj is vertically upwards. Then Horizontal range of the ground is ………… m

Velocity of a projectile at height 15 m from ground is v=(20hati+10hatj)m//s . Here hati is in horizontal direction and hatj is vertically upwards. Then Horizontal range of the ground is ………… m

Assertion At height 20 m from ground , velocity of a projectile is v = (20 hati + 10 hatj) ms^(-1) . Here, hati is horizontal and hatj is vertical. Then, the particle is at the same height after 4s. Reason Maximum height of particle from ground is 40m (take, g = 10 ms^(-2))

Assertion At height 20 m from ground , velocity of a projectile is v = (20 hati + 10 hatj) ms^(-1) . Here, hati is horizontal and hatj is vertical. Then, the particle is at the same height after 4s. Reason Maximum height of particle from ground is 40m (take, g = 10 ms^(-2))

A particle is projected from gound At a height of 0.4 m from the ground, the velocity of a projective in vector form is vecv=(6hati+2hatj)m//s (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is (g=10m//s^(2))

A particle is projected from gound At a height of 0.4 m from the ground, the velocity of a projective in vector form is vecv=(6hati+2hatj)m//s (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is (g=10m//s^(2))