Answer to Question #100204 in Classical Mechanics for Beast

Question

The ball is thrown upright at a height of 1.4 meters. The height of the ball as a function of time is h (t) = 1.4 + 10t -4.9t2. Determine

a) the starting speed of the ball

b) the ascent time of the ball to the top of the ball

c) the speed at which the ball hits the ground

Answer 1

a) The velocity of the ball is the derivative of the height:

$$v(t)=h'(t)=10-4.9\cdot2t=10-9.8t$$

Hence the starting speed of the ball is:

$$v(0)=10-9.8\cdot0=10 m/s$$

b)The top (maximum height) is the point where the slope of the tangent line (derivative) is equal to zero:

$$10-9.8t=0$$ $$9.8t=10$$ $$t\approx1.02$$

c) The ball hits the ground when the height is equal to 0:

$$1.4+10t-4.9t^2 =0$$ $$49t^2 -100t-14=0$$ $$t_1 =\frac{100-\sqrt{10000+4\cdot 49\cdot 14}}{2\cdot 49}\approx -0.13$$ $$t_2 =\frac{100+\sqrt{10000-4\cdot 49\cdot 14}}{2\cdot 49}\approx2.2$$

Since t can't be negative, then the ball hits the ground after 2.2 seconds.

Hence the velocity at t=2.2 is:

$$v(2.2)=10-9.8\cdot2.2=10-21.56=-11.56$$

Speed is the magnitude of the velocity. Then the speed at which the ball hits the ground is 11.56 m/s.

Answer:

a) 10 m/s;

b) 2.2 s

c) 11.56 m/s