Question

An 5kg mass is thrown with an initial velocity of 15 m/s from a height of 10 meters, at an angle of 20 degrees wrt horizontal. Using the basic kinematic equations (x(f)=x(i)+v_i,xt+(1/2)a_xt², and v_f,x=v_i,xt+a_xt, with similar equations along y), determine how far it has moved in x- and y-, separately, after a time of 20 seconds, and also the x- and y-components of the final velocity after t seconds, assuming the only external force acting on the mass is gravity; from trigonometry, now determine the magnitude of velocity from v_f,x and v_f,y that you've just calculated..

Now check your answer for the final velocity magnitude you just calculated using a conservation of energy approach: PE_i+KE_i=PE_f+KE_f + heat, for the case where there is no energy lost to friction, e.g. (i.e., no wind resistance in this case and heat=0). Note the energy conservation approach considerably simplifies the calculation.

Answer 1

Initial Horizontal component of velocity = u_x = 15 cos(20) = 14.09 m/s

Initial vertical component of velocity = u_y = 15 sin(20) = 5.13 m/s

Horizontal acceleration a_x = 0 m/s²

Vertical acceleration a_y = g = -9.8 m/s² (take downward direction negative)

Take origin at point of launch ,x(i) = 0 and y(i) = o

Let's first find the time in which it will hit the ground

using y(t) = y(i)+  uy t + 1/2 a_y t²

y(t) = -10 m

-10 = 0 + 5.13 x t + 1/2 (-9.8) x t²

4.9 t² - 5.13 t - 10 = 0

Solve quadratic in t

Neglecting one negative value of t

We get T = 2 s

So. In 2 sec projected mass will hit the ground.

Horizontal distance travelled in 2 s, X(t=20s) = u_x x 2 = 14.09 x 2 = 28.18 m ( a_x = 0)

There is no horizontal acceleration, so horizontal velocity will remain constant throughout the flight

V_x(t) = 14.09 m/s

For Vertical Velocity

V_y(t) = u_y + at = 5.13 - 9.8t

Hence Net velocity at time t ,