Question

From the top of a building 78.4 m high, a ball is thrown horizontally which hits the ground at a distance. The line joining the top of the building to the point where the ball hits the ground makes an angle of 45° with the horizontal. Then initial velocity of the ball is

Answer 1

Answer:

From the top of a building 78.4 m high, a ball is thrown horizontally which hits the ground at a distance. The line joining the top of the building to the point where the ball hits the ground makes an angle of 45° with the horizontal. Then initial velocity of the ball is 1) 9.8 ms 2) 19.6 ms 3) 9.8 /2 ms 4) 19.6 /2 ms.

Answer 2

Diagram :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.4mm}\qbezier(0,5)(5,4.5)(5,0)\qbezier(0,5)(0,5)(5,0)\put(0,0){\line(0,1){5}}\put(-1,0){\line(1,0){7}}\put(0,5){\vector(1,0){3.5}}\multiput(-0.5,0)(0.5,0){13}{\qbezier(0,0)(0,0)(-0.3,-0.3)}\put(-1.8,2.5){\bf{78.4 m}}\put(4,4.9){ \bf{u = u_x} }\put(2.3,-1){\bf{R}}\qbezier(4.3,0)(4.2,0.3)(4.5,0.5)\put(3.5,0.3){ \bf{45^{\circ}} }\end{picture}$

★ First of all we need to find time taken by ball to reach at ground.

This question is based on concept of height to ground projectile motion.

➝ t = √2H/g

➝ t = √2(78.4)/9.8

➝ t = √2(8)

➝ t = √16

➝ t = 4 s

Now let's try to find range of projectile.

➝ tan 45° = H/R

➝ 1 = 78.4/R

➝ R = 78.4 m

★ Calculation of initial velocity :

➝ R = u × t

➝ 78.4 = u × 4

➝ u = 78.4/4

➝ u = 19.6 m/s