Question

A particle is projected with speed 10 M per second at angle 60 with the horizontal then the time at which its speed become half of initial <br />

Answer 1

okay so we have to split the initial velocity of the ball into 2 components to start with, horizontal and vertical. to do this we do velocity x sin(angle) for the vertical component, and velocity x cos(angle) for the horizontal component.

10 x sin(30) = 5 m/s

10 x cos(30) = 8.66 m/s

now assuming air resistance is ignored, the horizontal component of the projectile will remain constant, as there is no force acting on the object horizontally.

however gravity is acting on the ball vertically, accelerating it downwards. so we can use a suvat equation to find the velocity after 1 second.

u=5 v=? a=-9.81 t=1

v = u + at

v = 5 + (-9.81 x 1)

v = -4.81 m/s

so we know that after 1 second, the vertical component would of changed to -4.81m/s, or 4.81 m/s downwards. your questions asks for speed, so the negative can be ignored. now we can use Pythagoras, as we have to vectors acting at 90 degrees to each over, to find the combined velocity. so:

a^2 + b^2 = c^2

8.66^2 + 4.81^2 = c^2

98.13 = c^2

9.91m/s = c

enjoy :)

Answer 2

the correct answer is 9.91m/second2