Question

1) A baseball player wants to throw a snowball over a house which is 55 ft tall. If he is standing 75 from the house and throws the ball at an angle of 45 degrees, ET s, i 10.58 m/s 2.2ur

Answer 1

Hi 8. Given t Height of house, H 55 ft. Distance from the hour - 75 st 0= US = Angle of projectile Let u be the speed of the ball a HSS/ AS AF 75Jt - B - conversion of units & R 1 ft = 12 inch linch = 2.54cm = 0.00254 m lift - 0.3048m x 0.305m a) so Height of noun , H = SS ft - S5X0.305 = 16.775 m Max. Height of projectile, H = y²sin2o A 0-4 29 16-175 -40-45 . 2 x 9.8 u 2 = 16 775 X 2X 9.8 . g = 9.8 m/s² & sinus V2 U² - 328.7982 = 657.58 u = √65778 m/s u = 25.64 m/s - And Hence speed he need to throw the ball is 25.64 m/s tet Time taken for the ball to flylt T= ausino (total time taken for flight) Uring t-T/2) 24 25 6 u x sinus. - 2X28.64XL 9.8682) 81.28 9.8052) *2

5.232 T.41482 IT: 3.708]. And = 1.858 Hence time taken for the ball to Hy is 1.858 c For where ball hit the ground we have to find max, Range. We know that at o us. 1 R= UH R =u² sinzo 3 R= 4x55 R = 220ft. R = 220x 0.305 = 67.1m let velocity at which ball hit the ground be us 2 ug ucoso i- usinoj : черчSt-ushчо d) us= u(tai I uplo J.)+(4)2 + y = tumia I lug = 25.64 m/s An If He runs lofts & throw the ball let ve loft/ph V 10X0.305 = 3.05 m/s on starting u: 28.64 m/s Hence Relative motion of the body u lu-u 125.64-3.051 m/ Ur= 22159 m/s

Now on taking us be the new velocity We anume he throw ball again at us: Let di distance it hit the ground d = u² sin20 22-59x22.59 x singo' 9.8 da 510.3081 = 52.07m Hence distance where it hit the 52.07m. ground ry
In part b,please note that we have to find time for fly which is equal to 1/2(Total time of flight).

After solving this,time for fly comes out to be 1.85seconds.

To find the all the parts of the given question please remember all the mentioned formulae.