A projectile is thrown at an angle of 60 with the horizontal. After how much time will its inclination with the horizontal be 45 ? ( Given V= 147m/s) .
Answers
Answered by
23
let v velocity , when inclination goes to 45 with horizontal .
also velocity along x axis remains constant
so,
Ucos60=Vcos45
V=Ucos60/cos45=147cos60/cos45
hence Vy=Vsin45
=147cos60.tan45=147cos60
equation for y component
Vy=Uy+ayt
147cos60=147sin60-gt
t=(147sin60-147cos60)/g=147 (0.732)/20
=5.3802 sec
also velocity along x axis remains constant
so,
Ucos60=Vcos45
V=Ucos60/cos45=147cos60/cos45
hence Vy=Vsin45
=147cos60.tan45=147cos60
equation for y component
Vy=Uy+ayt
147cos60=147sin60-gt
t=(147sin60-147cos60)/g=147 (0.732)/20
=5.3802 sec
abhi178:
if you like please mark as brainliest
sorry answer is 5.38 sec
thanks i understand but last second step is wrong because g - 10
g-10 is what
step is not wrong please again check
147 (sin60-cos60)/g=147 (0.732)/2 x 10=5.38 sec
ooh thanks bhai
okay no mention it's my pleasure
Answered by
12
u=147m/s initial angle = 60°
u_x = 147 cos 60° m/s = 73.5 m/s
For the inclination of the velocity to be 45° with the horizontal, the vertical velocity u_y should be same as the horizontal velocity u_x, ie., 73.5 m/s
Apply the equation of motion with g = 10 m/s²
73.5 = 147 * sin 60° - 10 * t
t = 73.5 (√3 -1)/10 = 5.38 sec
u_x = 147 cos 60° m/s = 73.5 m/s
For the inclination of the velocity to be 45° with the horizontal, the vertical velocity u_y should be same as the horizontal velocity u_x, ie., 73.5 m/s
Apply the equation of motion with g = 10 m/s²
73.5 = 147 * sin 60° - 10 * t
t = 73.5 (√3 -1)/10 = 5.38 sec
am I right sir
thanks
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