Question

A particle is projected with velocity of 40√2 m/s making an
angle of 45° with the horizontal, the magnitude of velocity
of the particle at t = 1 s after projection, will be​

Answer 1

Answer:

50m/s

Explanation:

Given that the particle was launched with velocity 40 √2 in angle 45 with the horizontal, the initial vertical velocity will be

$40 \sqrt{2} \sin(45) m {s}^{ - 1} \\ = 40m {s}^{ - 1}$

and the initial horizontal velocity will be

$40 \sqrt{2} \cos(45) m {s}^{ - 1} \\ = 40m {s}^{ - 1}$

After 1 second, the vertical velocity will have reduced by the magnitude of g*1s, because gravity is working in opposite direction of the velocity.

Hence, after 1s, the vertical velocity will be

$40m {s}^{ - 1} - 10 m {s}^{ - 2} \times 1s \\ = 40m{ s}^{ - 1} - 10m {s}^{ - 1} \\ = 30m {s}^{ - 1}$

However, since gravity acts at 90° with the horizontal component of the velocity, it will be unaffected.

The resultant velocity will be calculated as

$\sqrt{ {30}^{2} + {40}^{2} } \\ = \sqrt{900 + 1600} \\ = \sqrt{2500} \\ = 50m {s}^{ - 1}$