Question

A projectile is fired with an initial speed of 250 m/s and angle of elevation 60. (recall g 9.8 m/s2. Round your answers to the nearest whole number.) (a) find the range of the projectile. (b) find the maximum height reached. (c) find the speed at impact.

Answer 1

Speed u= 250 m/s

Angle α=60°

g=9.8 m/s²

We have formula

Range R = u²sin2α/(2g)

R= (250² *sin 120)/(9.8*2)

R= 62500*√3/2*1/(9.8*2)

=15625√3/9.8

=2761 m

For maximum height

H= u²sin²α/2g

= 62500*3/4*1/(2*9.8)

=46875/(9.8*2)

=2391 m

Answer 2

Range = 2761m

Max height = 2391m

Speed of impact = 216m/s

Given:

Speed (u) = 250 m/s

Angle (α) =60°

g=9.8 m/s²

To Find:

(a) find the range of the projectile.

(b) find the maximum height reached.

(c) find the speed at impact.

Solution:

WKT;

• Range (R) = u²sin2α/(2g)

=> R= (250² *sin 120)/(9.8*2)

=> R= 62500*√3/2*1/(9.8*2)

=> 15625√3/9.8

Range = 2761 m

• Maximum Height (H) = u²sin²α/2g

=> H = 62500*3/4*1/(2*9.8)

=> H = 46875/(9.8*2)

Max Height = 2391 m

• Speed of impact (V) = $\sqrt{2gH}$

=> V = $\sqrt{2*9.8*2391}$

Speed of impact = 216 m/s

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