Question

A projectile is thrown horizontally from the top of a tower and strikes the ground after 3 second at an
angle of 45° with the horizontal. Find the height of the tower and speed with which the body was
projected. Given g = 9.8 m/s​

Answer 1

Answer

• Height = 44.1 m
• Initial speed of projectile = 29.4 m/s

Given

• A projectile is thrown horizontally from the top of a tower and strikes the ground after 3 second at an  angle of 45° with the horizontal [ g = 9.8 m/s² ]

To Find

• The height of the tower ( Sy )
• speed with which the body was  projected ( Ux )

Solution

Find attachment for clear vision of diagram

Use 2nd equation of motion for finding the height of tower

$\implies \tt s=ut+\dfrac{1}{2}at^2\\\\\implies \tt s_y=u_y t+\dfrac{1}{2}gt^2\\\\\implies \tt s_y=0(t)+\dfrac{1}{2}\times 9.8 \times 3^2[U_y=0\ , Since\ thrown\ horizontally]\\\\\implies \tt s_y=\dfrac{9.8 \times 9}{2}\\\\\implies \tt s_y=44.1\ m$

So , Height of the tower is 44.1 m

_______________________________

From attachment ,

$\tt \tan 45=\dfrac{v_y}{v_x}\\\\\tt \implies 1=\dfrac{v_y}{v_x}\\\\\tt \implies \tt v_x=v_y...(1)$

Use 1st equation of motion .

$\implies \tt v=u+at\\\\\implies \tt v_y=u_y+gt\\\\\implies \tt v_y=0+9.8 \times 3\\\\\implies \tt v_y=29.4\ m/s$

From (1) ,

$\tt v_y=v_x=29.4\ m/s$

Now , we need to find Ux .

So use 1st equation of motion .

$\implies \tt v_x=u_x+(0)t[Since,g_x=0]\\\\\implies \tt 29.4=u_x\\\\\implies \tt u_x=29.4\ m/s$