Question

The horizontal range,T and maximum height of projectile ,of initial velocity =49m/s with an angle 60degree with the horizontal
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Answer 1

Answer:

Explanation: IN urs question time is missing so I have calculate t=2s and g=10m/s'2

we know that, For horizontal range

                       x= ucos(theta)*t

                       x= 49m/s*cos60(degree)*2

                       x= 49*1/2*2m/s*S

cancelling 2 by 1/2 and also cancelling sec by 1/sec

      so,              x= 49m

for Time of flight

We Know that

                              T= 2usin(theta)/g

                              T= 2*49*sin60/10, m/s * 1/m/s'2         (sin60=root3/2)

                              T= 98*root3/2*1/10 s

                              T= 98*1.732*1/20 s                       (root3= 1.732)

                              T=169.736/20 s

                              T= 8.48s

now, Max height   Hmax = u'2sin'e(theta)/2g

                             Hmax =  (49)'2* sin'2(60)/2*10 , m'2/s'2 * 1/m/s'2

                            Hmax = 2401* (root3)'2/2'2 /20 m

                            Hmax= 2401 *3/4 *1/20 ,m

                           Hmax = 7203/80 ,m

                            Hmax = 90.03m