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A projectile is fired with velocity 80m/s at an angle of 60d...
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A projectile is fired with velocity 80m/s at an angle of 60degrees to the horizontal . Calculate the following

I)the maximum height reached by the ball at 4s

ii)the range

iii)the position of the ball from the ground after 10s

iv)the time to reach the maximum height

v)the least velocity of projection in order to achieve the same horizontal range and the angle of projection to achieve the same range when u=80m/s

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Ghana
04 October 2020

School not set Nigeria
05 October 2020
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I) maximum height = u2sin2θ / 2g

= 802sin260 / 2(10)

= 240m

Ii) Range = u2sin2θ/g

= 802sin2(60) / 10

= 554.26m

Iii) s = usinθt - 1/2 gt2

S = 80sin(60)(10) - 1/2 (10)(10)2

= 192.82 m

IV) t = usinθ/g

= 80sin(60)/10

= 6.93 seconds

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School not set Nigeria
05 October 2020
0

V) in order to achieve the same range, Rmax, θ = 45

RMax = umin2sin2(45)/g

Rmax = umin2/g

554.26 = umin2/10

Umin2 = 5542.6

Umin = 74.45m/s

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Ana,
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Umin means minimum or least velociy
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Ana,
Thanks
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