Question

From the top of a tower 49m high a ball is thrown with a velocity of 29.4m/s at elevation of 30° with the horizontal when where and with what velocity will it strike the ground​

Answer 1

Given:

Height of tower is 49 and a ball is thrown with a velocity of 29.4 m/s at an elevation of 30° with horizontal.

To find:

• Time , position and final velocity of the ball when it strikes the ground.

Calculation:

The Y component of Velocity of ball will be:

$\sf{u_{y}= u \sin( {30}^{ \circ} ) }$

$\sf{ = > u_{y}= 29.4 \times \dfrac{1}{2} }$

$\sf{ = > u_{y}= 14.7 \: m {s}^{ - 1} }$

So, let time taken to reach the ground be t ;

$\sf{ = > h = - u_{y} t + \dfrac{1}{2}g {t}^{2} }$

$\sf{ = > 49 = -14.7t + \dfrac{1}{2}(9.8) {t}^{2} }$

$\sf{ = > 49 = -14.7t + 4.9 {t}^{2} }$

$\sf{ = > 10= -3t+ {t}^{2} }$

$\sf{ = > {t}^{2} - 3t - 10 = 0 }$

$\sf{ = > (t - 5)(t + 2) = 0 }$

So, t = 5 seconds is the time after which the ball hits the ground.

Position at which the ball strikes be x distance away from bottom of tower.

$\sf{x = u_{x} \times t}$

$\sf{ = > x = v \cos( {30}^{ \circ} ) \times t}$

$\sf{ = > x = 29.4\cos( {30}^{ \circ} ) \times 5}$

$\boxed{ \sf{ = > x =127.3 \: m}}$

Velocity with which the ball strikes be v :

$\rm{v = \sqrt{ {(v_{x})}^{2} + {(v_{y})}^{2} } }$

$\rm{ = > v = \sqrt{ {(u_{x} )}^{2} + {(u_{y} + gt)}^{2} } }$

$\rm{ = > v = \sqrt{ {(25.46)}^{2} + {(14.7 + 50)}^{2} } }$

$\rm{ = > v = \sqrt{ {(25.46)}^{2} + {(64.7)}^{2} } }$

$\rm{ = > v = \sqrt{4834.3 } }$

$\boxed{ \rm{ = > \: v = 69.52 \: m {s}^{ - 1} }}$

Hope It Helps.

Answer 2

Explanation:

it is whole answer that's will be helps you