Question

1. A ball is thrown upward from the top of a 35-m tower, with an initial velocity of 80 m/s and an angle of 25 degree. a) Find the time to reach the ground and the distance R from the bottom of the tower to the point of impact in the figure below. b) Find the magnitude and direction of the velocity at the moment of impact.  

Answer 1

As per the given question,

Height of the tower (h)=35m

Initial velocity of the ball (v) = 80 m/s

Angle of projection $(\theta) = 25^\circ$

a) Time to reach the ground (t)=?

Distance from the bottom of the tower to the point of projection (R)=?

b) Magnitude and the direction of velocity, at the time of impact to the ground=?

Let the final velocity of the ball be $v_1$

Now, taking the vertical and horizontal component of initial velocity.

Vertical component of velocity along y axis,$(v_y)=v\sin25^\circ$

Velocity along the x axis $(v_x) = v\cos 25^\circ$

Now, applying the second law of motion,

$-h=ut-\frac{gt^2}{2}$

Now, substituting the values in the above equation,

$\Rightarrow -35 = v\sin 25^\circ t-\frac{gt^2}{2}$

$\Rightarrow -35 = 80\sin 25^\circ t-\frac{9.8 t^2}{2}$

$\Rightarrow 4.9 t^2-33.8 t-35=0$

comparing it with standard quadratic equation $ax^2+bx+c=0$

a=4.9, b =-33.8, c = -35

$\Rightarrow t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Now, substituting the values in the above,

$\Rightarrow t =\frac{-(-33.8)\pm \sqrt{(-33.8)^2+4\times 4.9\times(-35)}}{2\times4.9}$

$\Rightarrow t = \frac{33.8\pm \sqrt{1828.44}}{9.8}s$

$\Rightarrow t = \frac{33.8\pm 42.76}{9.8}s$

time can not be negative, hence we will consider positive value,

$\Rightarrow t =\frac{33.8+42.76}{9.8}s$

$\Rightarrow t = \frac{76.56}{9.8}s$

$\Rightarrow t = 7.81 s$

Range of the ball $(R)=v\cos 25^\circ\times t$

Now, substituting the values,

$R=80\cos 25^\circ \times 7.81 m$

$\Rightarrow R =566.26m$ from the the bottom of the tower to the point of impact.

b) Horizontal component of the velocity remain unchanged, because there is not any force in the horizontal direction, but the vertical component of the velocity will not constant, because there is gravity force in the vertical direction.

At the point of the impact to the ground, let the angle made by the ball is $\phi$

Hence $v_1 \sin\phi = v\sin 25^\circ - gt$

now substituting the values

$\Rightarrow v_1 \sin\phi = 80\sin 25^\circ - 9.8\times 7.81$

$\Rightarrow v_1 \sin\phi =-42.72....(i)$

the horizontal component velocity remain unchanged hence,

$v_1 \cos\phi = v\cos 25^\circ$

Now substituting the values,

$v_1 \cos\phi = 80\cos 25^\circ$

$\Rightarrow v_1\cos\phi = 72.50 ....(ii)$

from equation (i) and (ii)

$\tan\phi= \frac{-42.72}{72.50}$

$\Rightarrow \tan\phi = -0.58$

Hence, $\phi = -30.11^\circ$

and final velocity, $v_1 = \frac{72.5}{\cos (-30.11)}$

$v_1= 83.80 m/s$