Question

A projectile is fired at an angle of30 to the horizontal with a velocity of 40ms-1.calculate the velocity attained after 1s.(g=10ms^2

Answer 1

Answer:

Distance covered/Horizontal Range,R=U²Sin2A/g

Explanation:

[A stands for theta in degrees°]

U=40m/s, A =30°,g=10m/s

R=(40²×Sin 2×30)/10

R=(1600×Sin60)/10

R=1385.6/10

R=138.6m~139m

V²=U²+2aS

[V=40,a=?,S=139]

40²=0²+2×a×139

1600=0+278a

1600=278a

a=5.75m/s²

The velocity attained after one second

V=U+at

[t=1s,a=5.75,U=?,V=40]

40=U+5.75(1)

40=U+5.75

U=40-5.75

U=34.25m/s