Question

A boy can throw a stone to maximum height of 50 m. To what maximum range can he throw this stone and to what height so that the maximum range is maintained?

• A. 100 m, 100 m
• B. 100 m, 25 m
• C. 200 m, 50 m
• D. 100 m, 50 m

Answer 1

Answer: (B)

100 m, 25 m

Answer 2

For maximum height, the boy has to throw the stone at θ = 90^o, then

H = u²/2g or u² = 2gH

or u² = 2 × 9.8 × 50 = 980 (ms^-1)²

maximum range R = u²/g = 980/9.8 = 100 m

Greatest height for maximum range is given by

H = $\frac { \mathrm { u } ^ { 2 } } { 2 \mathrm {~g} } \sin ^ { 2 } 45 = \frac { \mathrm { u } ^ { 2 } } { 2 \mathrm {~g} } \left( \frac { 1 } { \sqrt { 2 } } \right) ^ { 2 } = \frac { \mathrm { u } ^ { 2 } } { 4 \mathrm {~g} }$.

= $\frac { 980 } { 4 \times 9.8 }$ = 25 M