(a,b,c) u T = 15 m/s (velocity of truck)

Range = 60 m

Range = distance travelled by truck u T × T

60 = 15 × T ⇒ T = 4s = Time of flight (of ball)

T = where u y = Vertical component of ball's vel. {wrt ground}

∴ = u y i.e, u y = = 20 m/s

Now vel. of truck = u x = horizontal component of ball's vel. (wrt ground)

⇒ u x = 15 m/s (wrt ground)

This is because both cover same horizontal distance in same time with constant velocity along horizontal.

Now, velocity, of vall wrt track = V BT Then V BTx = V Bx – V Tx

i.e., velocity ball wrt truck (along x axis) = velocity of ball (wrt earth, along x axis) – velocity of truck (along x axis)

∴ V BT x = 15 – 15 = 0

Similarly, V BT y = V By – V Ty = 20 – 0

V BTy = 20 m/s.

⇒ = + = 0 + 20 m/s

⇒ velocity of ball wrt truck = 20 m/s upwards

velocity of ball, = + = 15 + 20

tan θ = θ = 53º

∴ speed = = = = 25 m/s

i.e., vel. of ball (wrt ground) = 25 m/s at an angle of 53º with the horizontal (as shown)